These notes accompany the Supersymmetry course held in the Epiphany term of the 2021-2022 academic year as part of the Particles, Strings and Cosmology MSc degree at Durham University.

Please send comments and corrections to Ben Hoare at ben.hoare-at-durham.ac.uk.

Durham, 10 January 2022

last updated 22 March 2022

This course provides a basic introduction to the main motivations and ideas in Supersymmetry (SUSY).

**Outline.**

What is SUSY and why is it interesting?

Poincaré and Lorentz groups, and spinor representations.

Supersymmetry algebra and on-shell representations (supermultiplets).

4-dimensional \(\mathcal{N}=1\) superspace and superfields.

SUSY field theories of chiral superfields. Holomorphy and non-renormalization theorems.

Vector superfields and supersymmetric gauge theories (SQCD, MSSM).

Spontaneous SUSY breaking.

**Structure.**

Details about the structure of the course can be found on Blackboard Ultra at

`https://blackboard.durham.ac.uk/ultra/courses/_18033_1/outline`

**Literature.**

There is a wealth of resources on supersymmetry freely available on the internet, along with a number of well-respected books on the subject that should be readily available in academic libraries. If you find that something is not completely clear to you, or you would like to explore certain topics further, in addition to asking questions in the surgeries and tutorials, you are strongly encouraged to consult alternative sources. Listed below is a range of lecture notes, books and reviews covering the various topics that will be discussed in the lectures.

**Books:**

Supersymmetry and Supergravity,

J. Wess and J. Bagger (Princeton University Press (1992));The Quantum Theory of Fields Volume 3: Supersymmetry,

S. Weinberg (Cambridge University Press (2000));Introduction to Supersymmetry and Supergravity,

P. West (World Scientific (1990));Modern Supersymmetry: Dynamics and Duality,

J. Terning (Oxford University Press (2006)).

**Reviews and lecture notes:**

Introduction to Supersymmetry,

R. Argurio,`http://homepages.ulb.ac.be/~rargurio/susycourse.pdf`

;An Introduction to Global Supersymmetry,

P. Argyres,`https://homepages.uc.edu/~argyrepc/cu661-gr-SUSY/index.html`

;Lectures on Supersymmetry,

M. Bertolini`https://people.sissa.it/~bertmat/susycourse.pdf`

;Introduction to Supersymmetry,

A. Bilal,`arXiv:hep-th/0101055`

;BUSSTEPP Lectures on Supersymmetry,

J. Figueroa-O’Farrill,`arXiv:hep-th/0109172`

;Lectures on supersymmetric gauge theories and electric-magnetic duality,

K. A. Intriligator and N. Seiberg,`arXiv:hep-th/9509066`

.A Supersymmetry Primer,

S. P. Martin,`arXiv:hep-ph/9709356`

;Introducing Supersymmetry,

M. F. Sohnius,**Phys. Rep. 128 (1985) 39-204**;An Unorthodox Introduction to Supersymmetric Gauge Theory,

M. J. Strassler,`arXiv:hep-th/0309149`

;Supersymmetric Field Theory,

D. Tong,`http://www.damtp.cam.ac.uk/user/tong/susy/susy.pdf`

.

Let us start with an overview of the key ideas behind supersymmetry and why it is of interest to both physicists and mathematicians. In this course we will then explore these ideas, develop the underlying mathematical formalism and investigate their physical implications. Our focus will be supersymmetry in quantum field theory (QFT) on 4-dimensional Minkowski space-time.

Supersymmetry is a space-time symmetry that maps fields or particles of integer spin to fields of particles of half-integer spin and vice versa. By the spin-statistics theorem, we know that fields of integer spin are bosonic, while those of half-integer spin are fermionic. Therefore, supersymmetry maps bosons to fermions and vice versa. Just as the familiar Poincaré symmetry of space-time (Lorentz transformations and space-time translations) give rise to conserved charges (angular momentum and momentum), the supersymmetries also give rise to conserved charges, known as *supercharges*. They are often denoted by \(Q\) and in the quantum theory map bosonic states to fermionic states and vice versa \[Q|\text{Boson}\rangle = |\text{Fermion}\rangle ~, \qquad Q |\text{Fermion}\rangle = |\text{Boson}\rangle ~.\] Conservation of charge immediately implies that the supercharges themselves must have half-integer spin. Fields or particles that are related by supersymmetry are known as *superpartners* and they sit in *supermultiplets*.

Given that the supercharges have non-zero spin, it follows that the supersymmetry algebra must non-trivially mix the Poincaré algebra with the supercharges. Indeed, the structure of the supersymmetry algebra takes the form \[\phantom{}[P,Q] = 0 ~, \qquad [M,Q] \sim Q ~, \qquad [B,Q] = 0 ~, \qquad [R,Q] \sim Q ~,\] where \(P\), \(M\) and \(B\) represent the generators of space-time translations, Lorentz transformations and internal (global and gauge) symmetries respectively. \(R\) represents the generators of the R-symmetry, a new type of symmetry that rotates the supercharges amongst themselves. From these commutation relations we can infer that two fields or particles that are superpartners will have the same mass and internal quantum numbers, but different spin and R-charge, as expected.

Since the supercharges have half-integer spin we expect them to have fermionic statistics. Therefore, to close the supersymmetry algebra, we also need to know how the supercharges *anticommute* amongst themselves \[\{Q,\bar Q\} \sim P ~.\] That is, schematically, we have that the square of a supersymmetry transformation is a space-time translation. It follows that if we gauge supersymmetry we also need to gauge the Poincaré symmetry and we are led to are theory known as supergravity: a gauge theory of local supersymmetry, which includes gravity.

It is possible to extend the Poincaré symmetry of QFT by different amounts of supersymmetry. The amount of supersymmetry is enumerated by an index \(I=1,\dots,\mathcal{N}\) carried by the supercharge \(Q^I\). The number of fields or particles in a supermultiplet grows exponentially with \(\mathcal{N}\), while the maximum spin grows linearly. A supersymmetric quantum field theory (SQFT) has no interacting degrees of freedom with spin greater than one, in particular they do not describe gravity. This constrains \(\mathcal{N} \leq 4\). This can be relaxed if we consider theories of supergravity, in which case there should be no interacting degrees of freedom with spin greater than two, and we have \(\mathcal{N} \leq 8\).

In this course we will focus on Lagrangian SQFTs on 4-dimensional Minkowski space-time. Furthermore, we will always consider theories that are defined by a set of fields and an action functional constrained by supersymmetry. There are SQFTs that do not have a Lagrangian formulation, but these are beyond the scope of this course.

There are many reasons why supersymmetry is of interest. Here we will focus on the reasons to study SQFT on 4-dimensional Minkowski space; however, generalisations to different dimensions and space-times, as well as to other areas of physics are also of great interest.

**Theoretical reasons.**

Supersymmetry is the most general symmetry of an interacting QFT. The Coleman-Mandula theorem (1967) states that in a unitary relativistic QFT with finitely many particles of mass less than \(M\) (for any finite \(M\)) and an analytic S-matrix, the Lie group of symmetries of the S-matrix is locally \[\textrm{Poincaré} \times \textrm{Compact internal symmetry} ~.\] A way to circumvent this theorem was understood by Golfand and Likhtman in 1971, which was generalised in 1975 by Haag, Łopuszański and Sohnius. The basic idea was to ask what happens if the symmetry algebra is extended to a superalgebra. It turns out that the maximal Lie supergroup of symmetries is locally \[\textrm{super-Poincaré} \times \textrm{Compact internal symmetry} \ltimes \textrm{R-symmetry} ~.\] Note that in both cases, relaxing the requirement of particle finiteness allows one to extend the Poincaré and super-Poincaré groups to the conformal and superconformal groups respectively.

Supersymmetry is a generic prediction of string theory. String theory, whether or not it ends up describing our universe, is an incredibly rich subject that has taught us a great deal about physics, including the nature of quantum gravity and how to approach solving strongly-coupled QFTs. It turns out that supersymmetry has many interesting consequences in string theory, and, in particular, is often taken to be crucial since it ensures the stability of the vacuum.

SQFTs are important theoretical laboratories. As we will see, supersymmetry gives rise to improved quantum behaviour and better theoretical control of QFTs. This can lead to exact results that hold both at weak and strong coupling, and allows us to explore the non-perturbative dynamics of gauge theories. It can also be used to probe the general structure of QFTs and has even had direct applications to areas of mathematics.

**Phenomenological reasons.**

Supersymmetry has important implications for the hierarchy problem. The electroweak scale \(M_{\text{EW}} \sim 10^2~\text{GeV}\), which is the symmetry-breaking scale of the Standard Model, is many orders of magnitude less than the Planck scale \(M_{\text{P}} \sim 10^{19}~\text{GeV}\). The hierarchy problem consists of two parts. First, why is \(M_{\text{EW}} \ll M_{\text{P}}\)? And second, is this hierarchy stable under quantum corrections? In the Standard Model the two-point function of the Higgs field receives one-loop corrections

coming from the Yukawa couplings \(\lambda H \bar\Psi\Psi\), where \(\Lambda_{\text{UV}}\) is the UV cut-off. Therefore, without any fine-tuning, loop corrections will typically lead to a Higgs mass of the same order as the scale at which new UV physics becomes relevant. This is often taken to be the Planck scale; however, the Higgs mass is many orders of magnitude less than the Planck scale. Therefore, this hierarchy is considered to be “unnatural” and requires fine-tuning. If there is also a scalar in the theory with coupling \(\lambda^2 H^2 \Phi^2\) then the two-point function of the Higgs field receives additional contributions at one loop

and the quadratic divergences cancel. Such cancellations often arise in SQFTs as a consequence of supersymmetry. If supersymmetry is broken at a scale close to \(M_{\text{EW}}\) then it should protect the Higgs mass from becoming too large.

Supersymmetry helps to construct grand unified theories (GUTs). One way to reduce the number of free parameters in the Standard Model is to conjecture that the \(\mathrm{SU}(3)\), \(\mathrm{SU}(2)\) and \(\mathrm{U}(1)\) gauge symmetries unify at higher energies into a single larger gauge group, such as \(\mathrm{SU}(5)\) or \(\mathrm{SO}(10)\). This symmetry should then be broken at the so-called GUT scale to the gauge group of the Standard Model \[\mathrm{G}_{\text{GUT}} ~ \xrightarrow[M_{\text{GUT}} \sim 10^{15}/10^{16}~\text{GeV}]{} ~ \mathrm{SU}(3) \times \mathrm{SU}(2) \times \mathrm{U}(1) ~ \xrightarrow[M_{\text{EW}} \sim 10^{2}~\text{GeV}]{} ~ \mathrm{SU}(3) \times \mathrm{U}(1)_{\text{EM}} ~.\] While the representations work nicely, there are a number of problems with this proposal. The three gauge couplings of the Standard Model don’t quite meet at the GUT scale. Yukawa couplings in the GUT induce proton decay; hence lower bounds on the proton lifetime constrain the GUT scale. Finally, there is a new hierarchy problem: why is \(M_{\text{EW}} \ll M_{\text{GUT}}\)? Introducing supersymmetry with low-energy supersymmetry breaking allows GUTs to be constructed without these problems.

Supersymmetry leads to natural dark matter candidates. Dark matter, matter that does not appear to interact with the electromagnetic field, is thought to make up around 25% of the mass-energy density of the universe. Many supersymmetric models of phenomenological interest contain a particle, the lightest superpartner (LSP), which is a natural candidate for dark matter. In particular, in these setups it is typically the case that an additional discrete symmetry known as R-parity implies that the LSP is a neutral particle that cannot decay to Standard Model particles; hence is stable.

Phenomenological reasons for studying supersymmetry may be becoming less relevant as data from the LHC rules out many models with low-energy supersymmetry breaking. Nevertheless, it is important to appreciate them since they motivated the investigation of SQFTs for many years, leading to significant advances in our theoretical understanding of QFT.

There are many more reasons why supersymmetry is of interest, and the brief explanations given here do not do justice to those we have stated. In this course you will be introduced to the basics of supersymmetry, with the aim that you acquire the skills needed to explore this fascinating subject in more depth.

Our goal is to construct quantum field theories with symmetries that map bosons to fermions and vice versa – supersymmetries. By the spin-statistics theorem we know that bosonic fields have integer spin, while fermionic fields have half-integer spin. It follows that to understand supersymmetry we will need to ensure that we understand the representation theory of the Lorentz and Poincaré groups. This is a beautiful story with many important implications for particle physics, which has Wigner’s classification of the (projective) irreducible unitary representations of the Poincaré group at its core.

We start from the definition of the Poincaré group as the isometry group of \(\mathbb{R}^{1,3}\), 4-dimensional Minkowski space-time. The metric is \[\begin{equation} \label{eq:metric} ds^2 = \eta_{\mu\nu} dx^\mu dx^\nu ~, \qquad \eta_{\mu\nu} = \mathop{\mathrm{diag}}(+1,-1,-1,-1)_{\mu\nu} ~. \end{equation}\] where \(x^\mu\) are coordinates on space-time and the index \(\mu=0,1,2,3\) is raised, lowered and contracted with \(\eta_{\mu\nu}\) and its inverse \(\eta^{\mu\nu}\). The isometries act on the coordinates as \[\begin{equation} \label{eq:isomtrans} x^\mu \to \Lambda^\mu{}_\nu x^\nu + c^\mu ~, \qquad \Lambda^\mu{}_\rho \Lambda^\nu{}_\sigma \eta_{\mu\nu} = \eta_{\rho\sigma} ~, \end{equation}\] where \(\Lambda^\mu{}_\nu \in \mathbb{R}\) and \(c^\mu \in \mathbb{R}\) are constant, i.e. independent of \(x^\mu\).

Check that the transformations \[x^\mu \to \Lambda^\mu{}_\nu x^\nu + c^\mu ~, \qquad \Lambda^\mu{}_\rho \Lambda^\nu{}_\sigma \eta_{\mu\nu} = \eta_{\rho\sigma} ~,\] are indeed isometries of the metric \[ds^2 = \eta_{\mu\nu} dx^\mu dx^\nu ~, \qquad \eta_{\mu\nu} = \mathop{\mathrm{diag}}(+1,-1,-1,-1)_{\mu\nu} ~,\] and that they form a group.

We have that \[\begin{split} x^\mu \to \Lambda^\mu{}_\nu x^\nu + c^\mu \qquad & \Rightarrow \qquad dx^\mu \to \Lambda^\mu{}_\nu dx^\nu \\ & \Rightarrow \qquad ds^2 = \eta_{\mu\nu} dx^\mu dx^\nu \to \eta_{\mu\nu} \Lambda^\mu{}_\rho \Lambda^\nu{}_\sigma dx^\rho dx^\sigma ~. \end{split}\] Therefore, if \(\Lambda^\mu{}_\rho \Lambda^\nu{}_\sigma \eta_{\mu\nu} = \eta_{\rho\sigma}\) we immediately see that the metric is invariant.

To check that the transformations form a group let us write \((\Lambda,c) x^\mu = \Lambda^\mu{}_\nu x^\nu + c^\mu\). We then have that \[(\tilde\Lambda,\tilde c)(\Lambda,c) x^\mu = \tilde\Lambda^\mu{}_\nu (\Lambda^\nu{}_\rho x^\rho + c^\nu) + \tilde c^\mu = (\tilde\Lambda\Lambda, \tilde\Lambda c + \tilde c) x^\mu ~.\] Given that \(\Lambda^\mathrm{t}\eta \Lambda = \eta\) and \(\tilde\Lambda^\mathrm{t}\eta\tilde\Lambda = \eta\) implies that the same is true for \(\tilde\Lambda\Lambda\), we find that the transformations close amongst themselves. The identity element is simply given by \((\Lambda,c) = (\mathbf{1},0)\) and the inverse of \((\Lambda,c)\) is \((\Lambda,c)^{-1} = (\eta\Lambda^\mathrm{t}\eta,-\eta\Lambda^\mathrm{t}\eta c)\). It is straightforward to check that these are of the required form using that \(\eta^{-1} = \eta^\mathrm{t}= \eta\). Finally, we need to check associativity. We have \[\begin{split} (\hat\Lambda,\hat c) \big((\tilde\Lambda,\tilde c)(\Lambda,c)\big) & = (\hat\Lambda,\hat c)(\tilde\Lambda\Lambda, \tilde\Lambda c + \tilde c) = (\hat\Lambda\tilde\Lambda\Lambda, \hat\Lambda\tilde\Lambda c + \hat\Lambda\tilde c + \hat c) ~, \\ \big((\hat\Lambda,\hat c)(\tilde\Lambda,\tilde c)\big)(\Lambda,c) & = (\hat\Lambda\tilde\Lambda, \hat\Lambda\tilde c + \hat c)(\Lambda,c) = (\hat\Lambda\tilde\Lambda\Lambda, \hat\Lambda\tilde\Lambda c + \hat\Lambda\tilde c + \hat c) ~, \end{split}\] showing that the transformations indeed form a group.

The transformations \(x^\mu \to \Lambda^\mu{}_\nu x^\nu\) generate the Lorentz group \(\mathrm{O}(1,3)\). The space-time translations \(x^\mu \to x^\mu + c^\mu\) generate a four-dimensional abelian group. Together these form the Poincaré group, which we denote \(\mathrm{IO}(1,3)\). The groups \(\mathrm{O}(1,3)\) and \(\mathrm{IO}(1,3)\) have four connected components, which are characterised by the sign of \(\Lambda^0{}_0\) and \(\det \Lambda = \pm 1\). The components with \(\Lambda^0{}_0 > 0\) and \(\det \Lambda = +1\) form subgroups known as the proper orthochronous Lorentz and Poincaré groups, which we denote \(\mathrm{SO}^+(1,3)\) and \(\mathrm{ISO}^+(1,3)\). The remaining three components are related to this component by time-reversal and parity. From now on we will drop the adjectives “proper orthochronous” and we will always mean the component connected to the identity when we write Lorentz or Poincaré group.

The Lorentz group \[\mathrm{SO}^+(1,3) = \Big\{\Lambda~:~\Lambda^\mu{}_\nu \in \mathbb{R},~\Lambda^\mathrm{t}\eta \Lambda = \eta,~\Lambda^0{}_0 > 0,~\det\Lambda = + 1\Big\} ~,\] is connected, but it is not simply connected. Its double (and universal) cover, denoted \(\mathrm{Spin}^+(1,3)\) is isomorphic to \(\mathrm{SL}(2,\mathbb{C})\). The group \(\mathrm{SL}(2,\mathbb{C})\) is defined as the set of \(2 \times 2\) complex matrices with determinant equal to 1 \[\begin{equation} \label{eq:sl2def} \mathrm{SL}(2,\mathbb{C}) = \Big\{N = \begin{pmatrix} a & b \\ c & d \end{pmatrix}~:~a,b,c,d\in\mathbb{C},~ad-bc = 1\Big\} ~. \end{equation}\] We would like to show that there is a 2 to 1 group homomorphism \[\mathrm{SL}(2,\mathbb{C}) \to \mathrm{SO}^+(1,3) ~.\] To specify this map we start by introducing the following two bases for the set of \(2 \times 2\) hermitian matrices \[\begin{equation} \label{eq:hermmat} \sigma_\mu = (\mathbf{1},\sigma_i) ~, \qquad \bar\sigma_\mu = (\mathbf{1},-\sigma_i) ~, \end{equation}\] where \(\sigma_i\), \(i=1,2,3\), are the usual Pauli matrices \[\sigma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} ~, \qquad \sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} ~, \qquad \sigma_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} ~.\] Note that \(\sigma_\mu = \bar \sigma^\mu\) and that the matrices \(\sigma_\mu\) and \(\bar\sigma^\mu\) satisfy \[\mathop{\mathrm{tr}}(\sigma_\mu \bar \sigma^\nu) = 2 \delta_{\mu}^{\nu} ~.\]

Given that we know how Lorentz transformations act on \(x^\mu\), one way to find the group homomorphism is to construct an action of \(\mathrm{SL}(2;\mathbb{C})\) on \(x^\mu\). For \(x^\mu \in \mathbb{R}^{1,3}\) we can construct the matrix \[X = x^\mu\sigma_\mu = \begin{pmatrix} x^0 + x^3 & x^1 - i x^2 \\ x^1 + i x^2 & x^0 - x^3 \end{pmatrix} ~,\] which is hermitian and satisfies \(\det X = x^\mu x_\mu\). Now consider the following linear transformation of \(X\) \[\begin{equation} \label{eq:sl2trans} X \to N X N^\dagger ~, \qquad N \in \mathrm{SL}(2,\mathbb{C}) ~. \end{equation}\] One can check that the set of such transformations forms a group. Furthermore, since \(\det N = \det N^\dagger = 1\), the determinant of \(X\) is preserved \[\det X \to \det X \qquad \Rightarrow \qquad x^\mu x_\mu \to x^\mu x_\mu ~.\] Therefore, a transformation of the form \(\eqref{eq:sl2trans}\) defines a Lorentz transformation of \(x^\mu\). Specifically, we have that \[\begin{equation} \label{eq:explicitmap} N x^\mu\sigma_\mu N^\dagger = \Lambda^\mu{}_\nu x^\nu \sigma_\mu \qquad \Rightarrow \qquad \Lambda^\mu{}_\nu = \frac12 \mathop{\mathrm{tr}}(N \sigma_\nu N^\dagger \bar \sigma^\mu) ~. \end{equation}\] This defines the group homomorphism \(\mathrm{SL}(2,\mathbb{C}) \to \mathrm{SO}^+(1,3)\). Note that, since \(\Lambda^0{}_0 = \frac12 \mathop{\mathrm{tr}}(N N^\dagger) = \frac12(|a|^2 + |b|^2 + |c|^2 + |d|^2) > 0\) (where \(N = \big(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\big)\)) and \(\det \Lambda = (\det N)^2(\det N^\dagger)^2 = +1\), we have that \(\Lambda \in \mathrm{SO}^+(1,3)\) as claimed.

Show that \[\Lambda^\mu{}_\nu = \frac12 \mathop{\mathrm{tr}}(N \sigma_\nu N^\dagger \bar \sigma^\mu) ~, \qquad N \in \mathrm{SL}(2,\mathbb{C}) ~,\] defines a group homomorphism \(\mathrm{SL}(2,\mathbb{C}) \to \mathrm{SO}^+(1,3)\).

We would like to show that \(\Lambda^\mu{}_\nu(N)\Lambda^\nu{}_\rho(\tilde{N}) = \Lambda^\mu{}_\rho(N\tilde{N})\). We have that \[\begin{split} \Lambda^\mu{}_\rho(N\tilde{N}) & = \frac{1}{2}\mathop{\mathrm{tr}}(N \tilde{N} \sigma_\rho \tilde{N}^\dagger N^\dagger \bar \sigma^\mu) \\ & = \frac{1}{4} \mathop{\mathrm{tr}}(\tilde{N}\sigma_\rho \tilde{N}^\dagger \bar\sigma^\nu ) \mathop{\mathrm{tr}}( \sigma_\nu N^\dagger \bar \sigma^\mu N) \\ & = \frac{1}{2} \mathop{\mathrm{tr}}(N \sigma_\nu N^\dagger \bar \sigma^\mu) \frac{1}{2} \mathop{\mathrm{tr}}(\tilde{N}\sigma_\rho \tilde{N}^\dagger \bar\sigma^\nu ) \\ & = \Lambda^\mu{}_\nu(N) \Lambda^\nu{}_\rho(\tilde{N}) ~, \end{split}\] as required. Here we have used the cyclicity of the trace, and, to go from the first line to the second line, that \(\sigma_\mu = \bar \sigma^\mu\) and that \(\{\tfrac{1}{\sqrt{2}}\sigma_\mu\}\) is an orthonormal basis (over \(\mathbb{C}\)) for the set of \(2 \times 2\) complex matrices; hence, \(\mathop{\mathrm{tr}}(M_1 M_2) = \frac12 \sum_\mu \mathop{\mathrm{tr}}(M_1 \sigma_\mu) \mathop{\mathrm{tr}}(\sigma_\mu M_2) = \frac12 \mathop{\mathrm{tr}}(M_1\bar\sigma^\mu) \mathop{\mathrm{tr}}(\sigma_\mu M_2)\).

Finally, we need to understand how the group of transformations \(\eqref{eq:sl2trans}\), which we have seen is isomorphic to \(\mathrm{SO}^+(1,3)\), is related to \(\mathrm{SL}(2,\mathbb{C})\). It is easy to show that there are exactly two choices of \(N \in \mathrm{SL}(2,\mathbb{C})\) generating each transformation of the type \(\eqref{eq:sl2trans}\), i.e. \(N\) and \(-N\). Therefore, the group homomorphism from \(\mathrm{SL}(2,\mathbb{C}) \to \mathrm{SO}^+(1,3)\) specified by \(\eqref{eq:explicitmap}\) is indeed 2 to 1 as claimed. Equivalently, we have that \(\mathrm{SO}^+(1,3)\) is isomorphic to \(\mathrm{SL}(2,\mathbb{C})/\mathbb{Z}_2\), where we quotient by the \(\mathbb{Z}_2\) centre of \(\mathrm{SL}(2,\mathbb{C})\) generated by \(\{\mathbf{1},-\mathbf{1}\}\).

We have seen that the Lorentz group \(\mathrm{SO}^+(1,3)\) is isomorphic to \(\mathrm{SL}(2,\mathbb{C})/\mathbb{Z}_2\). It follows that the corresponding Lie algebras \(\mathfrak{so}(1,3)\) and \(\mathfrak{sl}(2,\mathbb{C})\), which encode the infinitesimal transformations, are isomorphic. To define the Lie algebra \(\mathfrak{so}(1,3)\) we consider a curve \(\Lambda(t)\) in the Lorentz group with \(\Lambda(0) = \mathbf{1}\). By definition, we have that \(\Lambda(t)^\mathrm{t}\eta \Lambda(t) = \eta\). Differentiating this constraint with respect to \(t\) and setting \(t=0\) we find that \(M \in \mathfrak{so}(1,3)\) satisfies \[M^\mathrm{t}\eta + \eta M = 0 ~, \qquad M = \Big(\frac{d\Lambda(t)}{dt}\Big)\Bigg|_{t=0} ~.\] A basis for the set of \(4 \times 4\) matrices fulfilling this constraint is ^{1} \[\{M_{\mu\nu}~:~ (M_{\mu\nu})^\rho{}_\sigma = i(\eta_{\mu\sigma}\delta^\rho{}_\nu - \eta_{\nu\sigma}\delta^\rho{}_\mu) \} ~,\] which comprises of six linearly independent elements since \(M_{\mu\nu} = - M_{\nu\mu}\). An element of \(\mathfrak{so}(1,3)\) can then be written as \(M = \frac{i}{2} \omega^{\mu\nu} M_{\mu\nu}\) where \(\omega_{\mu\nu} = -\omega_{\mu\nu} \in\mathbb{R}\).

It is now straightforward to compute the Lie bracket of \(\mathfrak{so}(1,3)\), which takes the form \[\begin{equation} \label{eq:lorentz} \phantom{}[M_{\mu\nu},M_{\rho\sigma}] = - i \eta_{\mu\rho} M_{\nu\sigma} + i \eta_{\nu\rho} M_{\mu\sigma} - i \eta_{\nu\sigma} M_{\mu\rho} + i \eta_{\mu\sigma} M_{\nu\rho} ~. \end{equation}\] To show \(\mathfrak{so}(1,3)\) is isomorphic to \(\mathfrak{sl}(2,\mathbb{C})\) we define \[J_i = \frac{1}{2} \epsilon_{ijk} M_{jk} ~, \qquad K_i = M_{0i} ~,\] where the index \(i=1,2,3\) is contracted with \(\delta_{ij}\). It then follows that \[\phantom{}[J_i,J_j] = i \epsilon_{ijk} J_k ~, \qquad [K_i,K_j] = - i \epsilon_{ijk} J_k ~, \qquad [J_i,K_j] = i \epsilon_{ijk} K_k ~.\] Now setting \(A_i = J_i + i K_i\) we observe that in this basis a general element of \(\mathfrak{so}(1,3)\) is given by \(M = \tilde\omega_i A_i\) where \(\tilde\omega_i \in \mathbb{C}\), while the Lie bracket is \[\phantom{}[A_i,A_j] = i \epsilon_{ijk} A_k ~.\] Therefore, it is follows that \(M \in \mathfrak{sl}(2,\mathbb{C})\) and hence \(\mathfrak{so}(1,3)\) is isomorphic to \(\mathfrak{sl}(2,\mathbb{C})\).

To classify finite-dimensional irreducible representations of \(\mathfrak{so}(1,3)\), or equivalently \(\mathfrak{sl}(2,\mathbb{C})\) understood as a 6-dimensional real Lie algebra, we use that the real-linear representations of the latter are in one-to-one correspondence with complex-linear representations of its complexification \(\mathfrak{sl}(2,\mathbb{C}) \oplus \mathfrak{sl}(2,\mathbb{C})\). Therefore, the representations are labelled by a pair of numbers \((m,n)\), which we call spins, where the first entry denotes the representation of one of the two \(\mathfrak{sl}(2,\mathbb{C})\) algebras, and the second entry, the representation of the other. We use the familiar labelling with \(m\) either a non-negative integer or half-integer denoting the representation of dimension \(2m+1\), and similarly for \(n\). Therefore, the representation \((m,n)\) has dimension \((2m+1)(2n+1)\). Since \(\mathrm{SL}(2,\mathbb{C})\) is simply connected, these Lie algebra representations correspond to representations of the group; however, they may be projective representations of \(\mathrm{SO}^+(1,3)\).

Let us now describe a few of the lowest-dimensional representations. In the following we will always consider \(\mathfrak{sl}(2,\mathbb{C})\) and \(\mathrm{SL}(2,\mathbb{C})\) as a 6-dimensional real Lie algebra and group respectively.

**Spin 0.**

\((0,0)\): This is the 1-dimensional singlet representation, which acts on scalar fields \(\phi\).

**Spin \(\frac{1}{2}\).**

\((\frac{1}{2},0)\): This is one of the 2-dimensional complex representations, which acts on left-handed Weyl spinors \(\psi_\alpha\) with index \(\alpha = 1,2\). It is the fundamental representation of \(\mathrm{SL}(2,\mathbb{C})\), hence under Lorentz transformations \[\psi_\alpha \to N_\alpha{}^\beta \psi_\beta ~, \qquad \psi \to N \psi ~,\] where the \(2 \times 2\) matrix \(N\) is defined as in \(\eqref{eq:sl2def}\).

\((0,\frac{1}{2})\): This is the other 2-dimensional complex representation, which acts on right-handed Weyl spinors \(\bar\psi_{\dot\alpha}\) with index \(\dot{\alpha} = 1,2\). It is the conjugate of the fundamental representation of \(\mathrm{SL}(2,\mathbb{C})\), hence \[\bar\psi_{\dot{\alpha}} \to N^*{}_{\dot{\alpha}}{}^{\dot\beta} \bar \psi_{\dot\beta} ~, \qquad \bar \psi^\mathrm{t}\to \bar \psi^\mathrm{t}N^\dagger ~,\] where \(N^*\) is the complex conjugate of \(N\).

It follows that the complex conjugate of a left-handed spinor transforms as a right-handed spinor and vice versa. Therefore, we can always choose to write \(\bar\psi_{\dot\alpha} = (\psi_\alpha)^\dagger\).

Before we proceed to higher-dimensional representations, it will be useful to discuss the spinor representations in a bit more detail. To do so we introduce the following matrices \[\epsilon_{\alpha\beta} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}_{\alpha\beta} ~, \qquad
\epsilon_{\dot\alpha\dot\beta} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}_{\dot\alpha\dot\beta} ~,\] and their inverses \[\epsilon^{\alpha\beta} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}^{\alpha\beta} ~, \qquad
\epsilon^{\dot\alpha\dot\beta} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}^{\dot\alpha\dot\beta} ~,\] such that \(\epsilon^{\alpha\beta}\epsilon_{\beta\gamma} = \delta^\alpha_\gamma\) and \(\epsilon^{\dot\alpha\dot\beta}\epsilon_{\dot\beta\dot\gamma} = \delta^{\dot\alpha}_{\dot\gamma}\). The matrices \(\epsilon_{\alpha\beta}\) and \(\epsilon_{\dot\alpha\dot\beta}\) are invariant under the action of \(\mathrm{SL}(2,\mathbb{C})\). Indeed, we have ^{2} \[\epsilon_{\alpha\beta} \to N_\alpha{}^\gamma N_\beta{}^\delta \epsilon_{\gamma\delta} = \det N \epsilon_{\alpha\beta} = \epsilon_{\alpha\beta} ~,
\qquad
\epsilon \to N^\mathrm{t}\epsilon N = \epsilon ~,\] and a similar derivation holds for \(\epsilon_{\dot\alpha\dot\beta}\). This immediately allows us to infer that the dual of the fundamental representation is equivalent to the fundamental representation itself since \(N^{-\mathrm{t}} = \epsilon^{-1} N \epsilon\), i.e. \(N^{-\mathrm{t}}\) and \(N\) are related by a similarity transformation. Again, an analogous statement holds for the complex conjugate representations. Finally, we can show that the matrices \(\epsilon^{\alpha\beta}\) and \(\epsilon^{\dot\alpha\dot\beta}\) are also invariant under the action of \(\mathrm{SL}(2,\mathbb{C})\).

Given that \(\epsilon_{\alpha\beta}\) and \(\epsilon^{\alpha\beta}\), and \(\epsilon_{\dot\alpha\dot\beta}\) and \(\epsilon^{\dot\alpha\dot\beta}\), are invariant tensors, it is natural to use them to raise, lower and contract the indices \(\alpha = 1,2\) and \(\dot\alpha = 1,2\) respectively. Since this means that we are using antisymmetric tensors to raise, lower and contract indices, we fix our conventions to be ^{3} \[\psi^\alpha = \epsilon^{\alpha\beta} \psi_\beta ~, \qquad
\psi_\alpha = \epsilon_{\alpha\beta} \psi^\beta ~, \qquad
\bar\psi^{\dot\alpha} = \epsilon^{\dot\alpha\dot\beta} \bar\psi_{\dot\beta} ~, \qquad
\bar\psi_{\dot\alpha} = \epsilon_{\dot\alpha\dot\beta} \bar\psi^{\dot\beta} ~.\] Using these rules we can straightforwardly construct Lorentz scalars, e.g. \[\psi \chi \equiv \psi^\alpha\chi_\alpha ~,
\qquad
\bar \psi \bar \chi \equiv \bar\psi_{\dot\alpha}\bar\chi^{\dot\alpha} ~.\] Note that the indices are contracted differently in the two cases. Assuming \(\psi_\alpha\), \(\chi_\alpha\), \(\bar\psi_{\dot\alpha}\) and \(\bar\chi_{\dot\alpha}\) are anticommuting, which we will always assume to be the case, we have that \(\psi\chi = \chi\psi\) and \(\bar \psi \bar \chi = \bar \chi \bar \psi\). Furthermore, we use the convention \((\alpha\beta)^* = \beta^*\alpha^*\) for complex conjugation of anticommuting variables; hence \((\psi\chi)^\dagger = \bar\psi\bar\chi\).

A Dirac spinor is made up of one left-handed and one right-handed Weyl spinor, i.e. it is the reducible representation \((\frac12,0)\oplus(0,\frac12)\). To make this explicit, let us write a 4-component Dirac spinor as \[\Psi = \begin{pmatrix} \psi_\alpha \\ \bar\chi^{\dot\alpha} \end{pmatrix} ~.\] In this basis, known as the Weyl basis, the gamma matrices can be written in terms of the matrices \(\sigma_\mu\) and \(\bar\sigma_\mu\) \(\eqref{eq:hermmat}\). From the transformation rule \(\eqref{eq:sl2trans}\) we can see that \(\sigma_\mu\) should carry one undotted and one dotted index, in that order. Therefore, \[(\sigma_\mu)_{\alpha\dot\alpha} = (\mathbf{1},\sigma_i)_{\alpha\dot\alpha} ~, \qquad (\sigma^\mu)_{\alpha\dot\alpha} = (\mathbf{1},-\sigma_i)_{\alpha\dot\alpha} ~,\] while the matrices \(\bar\sigma_\mu\) are defined as \[(\bar\sigma_\mu)^{\dot\alpha\alpha} = \epsilon^{\dot\alpha\dot\beta}\epsilon^{\alpha\beta}(\sigma_\mu)_{\beta\dot\beta} = (\mathbf{1},-\sigma_i)^{\dot\alpha\alpha} ~, \qquad (\bar\sigma^\mu)^{\dot\alpha\alpha} = \epsilon^{\dot\alpha\dot\beta}\epsilon^{\alpha\beta}(\sigma^\mu)_{\beta\dot\beta} = (\mathbf{1},\sigma_i)_{\dot\alpha\alpha} ~.\] The gamma matrices can then be taken to be \[\gamma^\mu = \begin{pmatrix} \mathbf{0}& \sigma^\mu \\ \bar \sigma^\mu & \mathbf{0}\end{pmatrix} ~,\] which indeed satisfy the Clifford algebra relations \[\{\gamma^\mu,\gamma^\nu\} = 2 \eta^{\mu\nu} \mathbf{1}~,\] Computing \(\gamma_5 = i \gamma^0\gamma^1\gamma^2\gamma^3\) we find that \(\gamma_5 = \big(\begin{smallmatrix} \mathbf{1}& \mathbf{0}\\ \mathbf{0}& -\mathbf{1}\end{smallmatrix}\big)\) and we see that \(\big(\begin{smallmatrix}\psi_\alpha \\ \mathbf{0}\end{smallmatrix}\big)\) and \(\big(\begin{smallmatrix}\mathbf{0}\\ \bar\chi^{\dot\alpha}\end{smallmatrix}\big)\) are spinors of opposite chirality as expected.

Using the gamma matrices we can construct the generators of \(\mathfrak{sl}(2,\mathbb{C})\) in the fundamental representation and its conjugate. As usual we define \[\Sigma^{\mu\nu} = \frac{i}{4}[\gamma^\mu,\gamma^\nu] = i\begin{pmatrix} \sigma^{\mu\nu} & \mathbf{0}\\ \mathbf{0}& \bar\sigma^{\mu\nu}\end{pmatrix} ~.\] The block diagonal nature of these matrices highlights that we are dealing with a reducible representation, and it follows that \(i\sigma^{\mu\nu}\) and \(i\bar\sigma^{\mu\nu}\), where \[(\sigma^{\mu\nu})_{\alpha}{}^\beta = \frac{1}{4}(\sigma^\mu\bar\sigma^\nu - \sigma^\nu\bar\sigma^\mu)_{\alpha}{}^\beta ~, \qquad (\bar\sigma^{\mu\nu})^{\dot\alpha}{}_{\dot\beta} = \frac{1}{4}(\bar\sigma^\mu\sigma^\nu - \bar\sigma^\nu\sigma^\mu)^{\dot\alpha}{}_{\dot\beta} ~,\] are the generators the fundamental representation and its conjugate respectively.

Show that \[i(\sigma^{\mu\nu})_{\alpha}{}^\beta = \frac{i}{4}(\sigma^\mu\bar\sigma^\nu - \sigma^\nu\bar\sigma^\mu)_{\alpha}{}^\beta ~, \qquad i(\bar\sigma^{\mu\nu})^{\dot\alpha}{}_{\dot\beta} = \frac{i}{4}(\bar\sigma^\mu\sigma^\nu - \bar\sigma^\nu\sigma^\mu)^{\dot\alpha}{}_{\dot\beta} ~,\] satisfy the commutation relations of the Lorentz algebra.

Let us define \[\begin{split} J_i & = \frac{i}{2}\epsilon_{ijk} \sigma_{jk} = - \frac{i}{8}\epsilon_{ijk}[\sigma_j,\sigma_k] = \frac{1}{4} \epsilon_{ijk} \epsilon_{jkl}\sigma_l = \frac{1}{2} \sigma_i ~, \\ K_i & = i\sigma_{0i} = \frac{i}{4}( -\sigma_i - \sigma_i) = - \frac{i}{2} \sigma_i ~. \end{split}\] It is now straightforward to compute \[\begin{split} \phantom{}[J_i,J_j] & = \frac{1}{4}[\sigma_i,\sigma_j] = \frac{i}{2}\epsilon_{ijk}\sigma_k = i\epsilon_{ijk}J_k ~, \\ \phantom{}[K_i,K_j] & = -\frac{1}{4}[\sigma_i,\sigma_j] = -\frac{i}{2}\epsilon_{ijk}\sigma_k = -i\epsilon_{ijk}J_k ~, \\ \phantom{}[J_i,K_j] & = -\frac{i}{4}[\sigma_i,\sigma_j] = \frac{1}{2}\epsilon_{ijk}\sigma_k = i\epsilon_{ijk}K_k ~, \end{split}\] and we recover the commutation relations of the Lorentz algebra in one of its well-known forms as required. A similar computation follows for generators of the conjugate of the fundamental representation, with the only difference being that \(K_i = \frac{i}{2} \sigma_i\).

Prove the following identities for anticommuting spinors: \[\begin{aligned} (i): & \quad \psi^\alpha\psi^\beta = -\frac{1}{2}\epsilon^{\alpha\beta} \psi\psi ~, & (ii): & \quad \bar\psi^{\dot\alpha}\bar\psi^{\dot\beta} = \frac{1}{2}\epsilon^{\dot\alpha\dot\beta} \bar\psi\bar\psi ~, \\ (iii): & \quad (\theta\phi)(\theta\psi) = -\frac{1}{2}(\theta\theta)(\phi\psi) ~, & (iv): & \quad (\bar\theta\bar\phi)(\bar\theta\bar\psi) = -\frac{1}{2}(\bar\theta\bar\theta)(\bar\phi\bar\psi) ~, \\ (v): & \quad \chi\sigma^\mu \bar\psi = -\bar\psi \bar\sigma^\mu \chi ~, & (vi): & \quad \chi\sigma^\mu \bar\sigma^\nu \psi = \psi \sigma^\nu \bar\sigma^\mu \chi ~, \\ (vii): & \quad (\chi\sigma^\mu\bar\psi)^\dagger = \psi\sigma^\mu\bar\chi ~, & (viii): & \quad (\chi\sigma^\mu \bar\sigma^\nu \psi)^\dagger = \bar\psi\bar\sigma^\nu\sigma^\mu\bar\chi ~, \\ (ix): & \quad (\theta\psi)(\theta\sigma^\mu\bar\phi) = -\frac{1}{2}(\theta\theta)(\psi\sigma^\mu\bar\phi) ~, & (x): & \quad (\bar\theta\bar\psi)(\bar\theta\bar\sigma^\mu\phi) = -\frac{1}{2}(\bar\theta\bar\theta)(\bar\psi\bar\sigma^\mu\phi) ~, \\ (xi): & \quad (\phi\psi)(\bar\chi\bar\theta) = \frac{1}{2}(\phi\sigma^\mu\bar\chi)(\psi\sigma_\mu\bar\theta) ~, & (xii): & \quad (\theta\sigma^\mu\bar\theta)(\theta\sigma^\nu\bar\theta) = \frac{1}{2}(\theta\theta)(\bar\theta\bar \theta)\eta^{\mu\nu} ~. \end{aligned}\]

\((i)\): \(\psi^\alpha\psi^\beta\) is antisymmetric on interchange of \(\alpha\) and \(\beta\). Therefore, \(\psi^\alpha\psi^\beta \propto \epsilon^{\alpha\beta}\). To fix the proportionality constant we contract with \(\epsilon_{\alpha\beta}\). This gives \(\psi\psi = c \epsilon^{\alpha\beta}\epsilon_{\alpha\beta} = -2c\); hence \(\psi^\alpha\psi^\beta = -\frac{1}{2}\epsilon^{\alpha\beta}\psi\psi\).

\((ii)\): \(\bar\psi^{\dot\alpha}\bar\psi^{\dot\beta}\) is antisymmetric on interchange of \(\dot\alpha\) and \(\dot\beta\). Therefore, \(\bar\psi^{\dot\alpha}\bar\psi^{\dot\beta} \propto \epsilon^{\dot\alpha\dot\beta}\). To fix the proportionality constant we contract with \(\epsilon_{\dot\alpha\dot\beta}\). This gives \(-\bar\psi\bar\psi = c \epsilon^{\dot\alpha\dot\beta}\epsilon_{\dot\alpha\dot\beta} = -2c\); hence \(\bar\psi^{\dot\alpha}\bar\psi^{\dot\beta} = \frac{1}{2}\epsilon^{\dot\alpha\dot\beta}\bar\psi\bar\psi\).

\((iii)\): We have \[\begin{split} (\theta\phi)(\theta\psi) & = \epsilon_{\alpha\beta}\epsilon_{\gamma\delta} \theta^\alpha\phi^\beta\theta^\gamma\psi^\delta = - \epsilon_{\alpha\beta}\epsilon_{\gamma\delta}\theta^\alpha\theta^\gamma\phi^\beta\psi^\delta \\ & = \frac{1}{2} \epsilon_{\alpha\beta}\epsilon_{\gamma\delta}\epsilon^{\alpha\gamma} (\theta\theta)\phi^\beta\psi^\delta = - \frac{1}{2} \epsilon_{\beta\delta}(\theta\theta)\phi^\beta\psi^\delta = -\frac{1}{2}(\theta\theta)(\phi\psi) ~. \end{split}\]

\((iv)\): We have \[\begin{split} (\bar\theta\bar\phi)(\bar\theta\bar\psi) & = \epsilon_{\dot\alpha\dot\beta}\epsilon_{\dot\gamma\dot\delta} \bar\theta^{\dot\beta}\bar\phi^{\dot\alpha}\bar\theta^{\dot\delta}\bar\psi^{\dot\gamma} = -\epsilon_{\dot\alpha\dot\beta}\epsilon_{\dot\gamma\dot\delta} \bar\theta^{\dot\beta}\bar\theta^{\dot\delta}\bar\phi^{\dot\alpha}\bar\psi^{\dot\gamma} \\ & = -\frac{1}{2}\epsilon_{\dot\alpha\dot\beta}\epsilon_{\dot\gamma\dot\delta} \epsilon^{\dot\beta\dot\delta}(\bar\theta\bar\theta)\bar\phi^{\dot\alpha}\bar\psi^{\dot\gamma} = - \frac{1}{2} \epsilon_{\dot\gamma\dot\alpha}(\bar\theta\bar\theta)\bar\phi^{\dot\alpha}\bar\psi^{\dot\gamma} = -\frac{1}{2}(\bar\theta\bar\theta)(\bar\phi\bar\psi) ~. \end{split}\] Or alternatively, this is just the complex conjugate of identity \((iii)\).

\((v)\): We have \[\begin{split} \chi\sigma^\mu \bar\psi & = \chi^\alpha(\sigma^\mu)_{\alpha\dot\alpha} \bar\psi^{\dot\alpha} = \epsilon^{\alpha\beta}\chi_\beta (\sigma^\mu)_{\alpha\dot\alpha} \epsilon^{\dot\alpha\dot\beta}\bar\psi_\beta \\ & = \chi_\beta (\bar\sigma^\mu)^{\dot\beta\beta} \bar\psi_\beta = - \bar\psi_\beta(\bar\sigma^\mu)^{\dot\beta\beta}\chi_\beta = -\bar\psi \bar\sigma^\mu \chi ~. \end{split}\]

\((vi)\): We have \[\begin{split} \chi\sigma^\mu \bar\sigma^\nu \psi & = \chi^\alpha(\sigma^\mu)_{\alpha\dot\alpha}(\bar\sigma^\nu)^{\dot\alpha\beta}\psi_\beta = \epsilon^{\alpha\gamma}\chi_\gamma(\sigma^\mu)_{\alpha\dot\alpha}\epsilon^{\dot\alpha\dot\beta}\epsilon_{\dot\beta\dot\gamma} (\bar\sigma^\nu)^{\dot\gamma\beta} \epsilon_{\beta\delta}\psi^\delta \\ & = - \chi_\gamma(\bar\sigma^\mu)^{\dot\beta\gamma}(\sigma^\nu)_{\delta\dot\beta} \psi^\delta = \psi^\delta(\sigma^\nu)_{\delta\dot\beta}(\bar\sigma^\mu)^{\dot\beta\gamma}\chi_\gamma =\psi \sigma^\nu \bar\sigma^\mu \chi ~, \end{split}\]

\((vii)\): We have \[(\chi\sigma^\mu\bar\psi)^\dagger = \psi(\sigma^\mu)^\dagger \bar\chi = \psi\sigma^\mu\bar\chi ~.\]

\((viii)\): We have \[(\chi\sigma^\mu\bar\sigma^\nu\bar\psi)^\dagger = \psi(\bar\sigma^\nu)^\dagger(\sigma^\mu)^\dagger \bar\chi = \psi\bar\sigma^\nu\sigma^\mu\bar\chi ~.\]

\((ix)\): Follows from identity \((iii)\) viewing \(\sigma^\mu \bar\phi\) as a left-handed Weyl spinor with an undotted index.

\((x)\): Follows from identity \((iv)\) viewing \(\bar\sigma^\mu \phi\) as a right-handed Weyl spinor with a dotted index. Or alternatively, this is just the complex conjugate of identity \((ix)\).

\((xi)\): We have that \[(\phi\psi)(\bar\chi\bar\theta) = \phi^\alpha \psi_\alpha \bar\chi_{\dot\alpha} \bar\theta^{\dot\alpha} = -\bar\chi^{\dot\alpha}\phi^\alpha\psi_\alpha \bar\theta_{\dot\alpha}= - \mathop{\mathrm{tr}}(M_1 M_2) ~,\] where \((M_1)^{\dot\alpha\alpha} = \bar\chi^{\dot\alpha}\phi^\alpha\) and \((M_2)_{\alpha\dot\alpha} = \psi_\alpha\bar\theta_{\dot\alpha}\). Since \(\sigma^\mu = \bar\sigma_\mu\) and \(\{\tfrac{1}{\sqrt{2}}\sigma^\mu\}\) is an orthonormal basis (over \(\mathbb{C}\)) for the set of \(2\times 2\) complex matrices, we have that \(\mathop{\mathrm{tr}}(M_1M_2) = \frac{1}{2}\mathop{\mathrm{tr}}(M_1\sigma^\mu)\mathop{\mathrm{tr}}(M_2\bar\sigma_\mu)\). Substituting back in for \(M_1\) and \(M_2\) gives \[(\phi\psi)(\bar\chi\bar\theta) = - \frac{1}{2}(\phi\sigma^\mu\bar\chi)(\bar\theta\bar\sigma_\mu\psi) = \frac{1}{2}(\phi\sigma^\mu\bar\chi)(\psi\sigma_\mu\bar\theta) ~,\] where we have used identity \((v)\).

\((xii)\): We have \[\begin{split} (\theta\sigma^\mu\bar\theta)(\theta\sigma^\nu\bar\theta) & = - \theta^\alpha \theta^\beta \bar\theta^{\dot\alpha}\bar\theta^{\dot\beta} (\sigma^\mu)_{\alpha\dot\alpha} (\sigma^\nu)_{\beta\dot\beta} = \frac{1}{4} (\theta\theta)(\bar\theta\bar\theta) \epsilon^{\alpha\beta}\epsilon^{\dot\alpha\dot\beta} (\sigma^\mu)_{\alpha\dot\alpha} (\sigma^\nu)_{\beta\dot\beta} \\ & = \frac{1}{4} (\theta\theta)(\bar\theta\bar\theta) (\sigma^\mu)_{\alpha\dot\alpha} (\bar\sigma^\nu)^{\dot\alpha\alpha} = \frac{1}{4} (\theta\theta)(\bar\theta\bar\theta) \mathop{\mathrm{tr}}(\sigma^\mu\bar\sigma^\nu) =\frac{1}{2}(\theta\theta)(\bar\theta\bar\theta)\eta^{\mu\nu} ~. \end{split}\]

An alternative way to construct an irreducible representation from a Dirac spinor is to impose a reality condition, known as a Majorana condition, relating the left-handed and right-handed Weyl spinors. To do this we set \[\psi_\alpha = \chi_\alpha ~, \qquad \Psi_{\text{M}} = \begin{pmatrix} \psi_\alpha \\ \bar\psi^{\dot\alpha} \end{pmatrix} ~,\] where we recall that \(\chi_\alpha\) is defined as the complex conjugate of \(\bar \chi_{\dot\alpha}\). This reality condition is compatible with Lorentz symmetry since both \(\psi_\alpha\) and \(\chi_\alpha\) transform as left-handed Weyl spinors. Note that if we try to impose the Weyl condition on top of the Majorana condition then the spinor is forced to vanish. Indeed, in 4 dimensions, both the Majorana and Weyl spinors are irreducible representations and there is no Majorana-Weyl spinor.

A Dirac spinor is made up of one left-handed and one right-handed Weyl spinor. In the Weyl basis we have \[\Psi = \begin{pmatrix} \psi_\alpha \\ \bar\chi^{\dot\alpha} \end{pmatrix} ~, \qquad \gamma^\mu = \begin{pmatrix} \mathbf{0}& \sigma^\mu \\ \bar \sigma^\mu & \mathbf{0}\end{pmatrix} ~.\] Check that the gamma matrices \(\gamma^\mu\) satisfy the Clifford algebra relations \(\{\gamma^\mu,\gamma^\nu\} = 2\eta^{\mu\nu}\mathbf{1}\). Express the Dirac Lagrangian \[\mathcal{L}= -i\bar\Psi\gamma^\mu\partial_\mu \Psi - m \bar\Psi\Psi ~,\] in terms of the Weyl spinors \(\psi\) and \(\bar\chi\). What happens when \(\Psi\) is taken the be a Majorana spinor?

Computing the anticommutator of the gamma matrices we find \[\{\gamma^\mu,\gamma^\nu\} = \begin{pmatrix} \sigma^\mu\bar\sigma^\nu +\sigma^\nu\bar\sigma^\mu & \mathbf{0}\\ \mathbf{0}& \bar\sigma^\mu\sigma^\nu +\bar\sigma^\nu\sigma^\mu \end{pmatrix}~.\] We can then check explicitly that \[\begin{split} \sigma^0\bar\sigma^0 +\sigma^0\bar\sigma^0 & = 2 \mathbf{1}~, \\ \sigma^0\bar\sigma^i +\sigma^i\bar\sigma^0 & = \sigma_i - \sigma_i = 0 ~, \\ \sigma^i\bar\sigma^j +\sigma^j\bar\sigma^i & = -\{\sigma_i,\sigma_j\}= - 2\delta_{ij}\mathbf{1}~, \end{split}\] so that we have \(\sigma^\mu\bar\sigma^\nu +\sigma^\nu\bar\sigma^\mu= 2 \eta^{\mu\nu}\mathbf{1}\). Similarly, we find that \(\bar\sigma^\mu\sigma^\nu +\bar\sigma^\nu\sigma^\mu = 2 \eta^{\mu\nu}\mathbf{1}\) and we have that the gamma matrices indeed satisfy the Clifford algebra relations.

To express the Dirac Lagrangian in terms of Weyl spinors, let us start by recalling that \(\bar\Psi = \Psi^\dagger \gamma^0\). Therefore, \[\bar\Psi = \begin{pmatrix} \bar\psi_{\dot\alpha} & \chi^{\alpha} \end{pmatrix} \begin{pmatrix} \mathbf{0}& \mathbf{1}\\ \mathbf{1}& \mathbf{0}\end{pmatrix} = \begin{pmatrix} \chi^{\alpha} & \bar\psi_{\dot\alpha} \end{pmatrix} ~.\] Now substituting into the Dirac Lagrangian gives \[\begin{split} \mathcal{L}& = -i\bar\Psi\gamma^\mu\partial_\mu \Psi - m \bar\Psi\Psi \\ & = -i \begin{pmatrix} \chi^{\alpha} & \bar\psi_{\dot\alpha} \end{pmatrix} \begin{pmatrix} \mathbf{0}& (\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu \\ (\bar \sigma^\mu)^{\dot\alpha\alpha} \partial_\mu & \mathbf{0}\end{pmatrix} \begin{pmatrix} \psi_\alpha \\ \bar\chi^{\dot\alpha} \end{pmatrix} - m\begin{pmatrix} \chi_{\alpha} & \bar\psi^{\dot\alpha} \end{pmatrix} \begin{pmatrix} \psi_\alpha \\ \bar\chi^{\dot\alpha} \end{pmatrix} \\ & = - i\chi^{\alpha}(\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu\bar\chi^{\dot\alpha} -i\bar\psi_{\dot\alpha}(\bar\sigma^\mu)^{\dot\alpha\alpha} \partial_\mu \psi_\alpha -m \chi^{\alpha}\psi_\alpha -m \bar\psi_{\dot\alpha}\bar\chi^{\dot\alpha} \\ & = -i( \chi\sigma^\mu\partial_\mu\bar\chi +\bar\psi\bar\sigma^\mu \partial_\mu\psi) -m (\chi\psi + \bar\psi\bar\chi) \\ & = -i(\bar\chi\bar\sigma^\mu \partial_\mu\chi +\bar\psi\bar\sigma^\mu \partial_\mu\psi) -m(\chi\psi + \bar\psi\bar\chi) ~, \end{split}\] where to go to the final line we have integrated by parts and dropped a total derivative. For a Majorana spinor, we simply replace \(\chi\) by \(\psi\).

**Spin 1.**

\((\frac{1}{2},\frac{1}{2})\): This 4-dimensional representation is the final representation that we will discuss in detail. It acts on an object that carries one undotted and one dotted index \(X_{\alpha\dot\alpha}\). This is not irreducible as a complex representation. Indeed, we can impose the reality condition \(X = X^\dagger\), which is preserved by the action of the Lorentz group \[X \to N X N^\dagger ~.\] As we have already seen, if we write a hermitian \(2\times 2\) matrix transforming in this way as \(X = x^\mu \sigma_\mu\) then \(x^\mu\) transforms in the vector representation of \(\mathrm{SO}^+(1,3)\). Therefore, we have that the \((\frac{1}{2},\frac{1}{2})\) representation of \(\mathrm{SL}(2,\mathbb{C})\) is equivalent to the vector representation of \(\mathrm{SO}^+(1,3)\). Furthermore, the matrices \((\sigma^\mu)_{\alpha\dot\alpha}\) realise the isomorphism between the two.

In general, the representations \((m,n)\) (\(m \neq n\)) will be irreducible as complex representations, with \((n,m)\) the conjugate representation of \((m,n)\). The representations \((m,m)\) admit a reality condition to give real irreducible representations. Other real irreducible representations can be constructed by taking the direct sum \((m,n)\oplus(n,m)\) and imposing a reality condition relating the two complex irreducible representations.

Can you think of any fields that you have encountered that transform in the real irreducible representations \((1,0)\oplus(0,1)\) and \((1,1)\)?

The field strength \(F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu\) of a \(\mathrm{U}(1)\) gauge field transforms in the \((1,0)\oplus(0,1)\) representation. The two complex irreducible representations correspond to the self-dual and anti-self-dual part. As expected, we need to complexify the field strength to construct these since the Hodge dual squares to \(-1\) on \(\mathbb{R}^{1,3}\). The traceless part of a symmetric energy-momentum tensor \(T_{\mu\nu}\) transforms in the \((1,1)\) representation.

A complex second rank antisymmetric tensor \(f^{\mu\nu} = - f^{\nu\mu}\) can be decomposed into a self-dual part \(f_+^{\mu\nu}\) and an anti-self-dual part \(f_-^{\mu\nu}\) \[f^{\mu\nu} = f_+^{\mu\nu} + f_-^{\mu\nu} ~, \qquad f_\pm^{\mu\nu} = \frac12 \Big(f^{\mu\nu} \pm \frac{i}{2}\epsilon^{\mu\nu\rho\sigma}f_{\rho\sigma}\Big) ~,\] where \(\epsilon^{\mu\nu\rho\sigma}\) is the Levi-Civita symbol with \(\epsilon^{0123} = 1\). The self-dual and anti-self-dual parts are therefore constrained such that \[f_\pm^{\mu\nu} = \pm \frac{i}{2}\epsilon^{\mu\nu\rho\sigma}f_\pm{}_{\rho\sigma} ~.\] Check that \(\sigma^{\mu\nu} = \frac{1}{4}(\sigma^\mu\bar\sigma^\nu - \sigma^\nu\bar\sigma^\mu)\) and \(\bar\sigma^{\mu\nu} = \frac{1}{4}(\bar\sigma^\mu\sigma^\nu - \bar\sigma^\nu\sigma^\mu)\) are self-dual and anti-self-dual respectively. Raising and lowering indices with \(\epsilon_{\alpha\beta}\), \(\epsilon_{\dot\alpha\dot\beta}\) and their inverses, show that \((\sigma^{\mu\nu})_{\alpha\beta}\) and \((\bar\sigma^{\mu\nu})^{\dot\alpha\dot\beta}\) realise the isomorphism between self-dual and anti-self-dual second-rank antisymmetric tensors and the \((1,0)\) and \((0,1)\) representations of \(\mathrm{SL}(2,\mathbb{C})\).

We first note that \(\sigma^{\mu\nu}\) and \(\bar\sigma^{\mu\nu}\) are antisymmetric on interchange of \(\mu\) and \(\nu\). Next we substitute in for \(\sigma^\mu\) and \(\bar\sigma^\mu\) to find \[\sigma^{0i} = \frac{1}{2}\sigma_i ~, \qquad \bar\sigma^{0i} = -\frac{1}{2}\sigma_i ~, \qquad \sigma^{ij} = -\frac{i}{2}\epsilon_{ijk}\sigma_k ~, \qquad \bar\sigma^{ij} = -\frac{i}{2}\epsilon_{ijk}\sigma_k ~.\] Therefore, we have \[\begin{split} \frac{i}{2}\epsilon^{0ijk} \sigma_{jk} & = \frac{1}{4}\epsilon_{ijk}\epsilon_{jkl} \sigma_l = \frac{1}{2}\sigma_i = \sigma^{0i} ~, \\ \frac{i}{2}\epsilon^{ij\mu\nu}\sigma_{\mu\nu} & = i\epsilon^{ij0k}\sigma_{0k} = -\frac{i}{2}\epsilon_{ijk} \sigma_k = \sigma^{ij} ~, \end{split}\] confirming that \(\sigma^{\mu\nu} = \frac{i}{2}\epsilon^{\mu\nu\rho\sigma} \sigma_{\mu\nu}\). A similar computation holds for \(\bar\sigma^{\mu\nu}\).

The matrices \((\sigma^{\mu\nu})_{\alpha\beta} = \epsilon_{\beta\gamma} (\sigma^{\mu\nu})_{\alpha}{}^{\gamma}\) and \((\bar\sigma^{\mu\nu})^{\dot\alpha\dot\beta} = \epsilon^{\dot\beta\dot\gamma} (\bar\sigma^{\mu\nu})^{\dot\alpha}{}_{\dot\gamma}\) are symmetric on interchange of \(\alpha\) and \(\beta\) or \(\dot\alpha\) and \(\dot\beta\). This follows since \(\sigma_i\epsilon\) is symmetric for all the Pauli matrices \(\sigma_i\). Therefore, contracting \((\sigma^{\mu\nu})_{\alpha\beta}\) with a self-dual second-rank antisymmetric tensor gives a symmetric bispinor with two undotted indices, which corresponds to the irreducible representation \((1,0)\). Similarly, contracting \((\bar\sigma^{\mu\nu})^{\dot\alpha\dot\beta}\) with an anti-self-dual second-rank antisymmetric tensor gives a symmetric bispinor with two dotted indices, which corresponds to the irreducible representation \((0,1)\).

While understanding the representations of the Lorentz group and its double cover is useful for constructing Lagrangians, the physical states of our theory should transform in (projective) unitary irreducible representations of the Poincaré group \(\mathrm{ISO}^+(1,3)\). As for the Lorentz group, we are interested in representations of the universal cover of the Poincaré group, i.e. its double cover.

The Lie algebra of the Poincaré group is an extension of \(\mathfrak{so}(1,3)\), the Lie algebra of the Lorentz group. The new commutation relations, in addition to those of the Lorentz algebra \(\eqref{eq:lorentz}\), are \[\begin{equation} \label{eq:poincare} \phantom{}[P_\mu,P_\nu] = 0 ~, \qquad [M_{\mu\nu}, P_\rho] = - i \eta_{\mu\rho} P_\nu + i \eta_{\nu\rho}P_\mu ~. \end{equation}\] The Pauli-Lubański pseudovector is defined as \(W^\mu = \frac{1}{2}\epsilon^{\mu\nu\rho\sigma}P_\nu M_{\rho\sigma}\). There are then two Casimir invariants, which are given by \[P^2 = P^\mu P_\mu ~, \qquad W^2 = W^\mu W_\mu ~.\] This means that they commute with both \(M_{\mu\nu}\) and \(P_\mu\). Commuting with \(M_{\mu\nu}\) follows since they are both Lorentz scalars, while \(P^2\) trivially commutes with \(P_\mu\). Finally, one can show explicitly that \(W^2\) commutes with \(P^\mu\).

Show that \([W^2,P_\mu] = 0\) where \(W^\mu\) is the Pauli-Lubański vector \(W^\mu = \frac{1}{2}\epsilon^{\mu\nu\rho\sigma}P_\nu M_{\rho\sigma}\).

We start by showing that \[\begin{split} \phantom{}[W_\mu,P_\tau] & = \frac{1}{2}\epsilon^{\mu\nu\rho\sigma}P_\nu[M_{\rho\sigma},P_\tau] = -\frac{i}{2}\epsilon^{\mu\nu\rho\sigma}P_\nu(\eta_{\rho\tau} P_\sigma - \eta_{\sigma\tau}P_\rho) \\ & = i\epsilon^{\mu\nu\rho}{}_{\tau}P_\nu P_\rho = \frac{i}{2}\epsilon^{\mu\nu\rho}{}_{\tau}[P_\nu, P_\rho] = 0 ~. \end{split}\] It then immediately follows that \([W^2,P_\mu] = 0\).

Since the translation generators commute amongst themselves, they can be simultaneously diagonalised. Therefore, if \(V\) is our space of states, we can write \[V = \bigoplus_{p} V_p ~, \qquad V_p = \{|\psi\rangle\in V ~:~ P_\mu |\psi\rangle = p_\mu |\psi\rangle \} ~.\] We define the little group \(\mathrm{H}_p\) associated with \(V_p\) to be the subgroup of the Poincaré group that preserves \(V_p\) \[H_p = \{(\Lambda,c) \in \mathrm{ISO}^+(1,3) ~:~ |\psi\rangle \in V_p ~\Rightarrow~ (\Lambda,c)|\psi\rangle \in V_p\} ~.\] It follows that \((\Lambda,c)p = p\) for \((\Lambda,c)\in H_p\).

To analyse the structure of the little group we start by observing that, while the whole Poincaré group does not leave \(p_\mu\) invariant, it does leave \(p^2\) invariant. This follows from the fact that \(P^2\) is a Casimir invariant. Therefore, we can take \[p^2 = m^2 ~.\] to be fixed for an irreducible representation. There are a number of cases that should be considered in a complete analysis. Here we focus on the two that are the most physically relevant: positive-energy massive and massless particles.

**Positive-energy massive particles.**

In this case we have \(m^2 > 0\) and \(p^0 > 0\). There exists a Lorentz transformation such that \(p = \Lambda k\) where \[k_\mu = (m,0,0,0)_\mu ~.\] It is then straightforward to see that the little group \(\mathrm{H}_k = \mathrm{SO}(3)\). Since \(p = \Lambda k\) we have that \(\mathrm{H}_p = \Lambda \mathrm{H}_k \Lambda^{-1}\); hence we also have that \(\mathrm{H}_p = \mathrm{SO}(3)\).

Massive on-shell physical states therefore transform in irreducible representations labelled by their mass \(m\) and a representation of \(\mathfrak{so}(3)\), which is isomorphic to \(\mathfrak{su}(2)\). The representation of \(\mathfrak{so}(3) \cong \mathfrak{su}(2)\) is specified by an integer or half-integer \(s\), which is the spin of the particle. The second Casimir invariant is given in terms of the mass and spin as \[W^2 = -m^2s(s+1) ~.\]

**Positive-energy massless particles.**

In this case we have \(m^2 = 0\) and \(p^0 > 0\). There exists a Lorentz transformation such that \(p = \Lambda k\) where \[k_\mu = (E,0,0,E)_\mu ~.\] One can then show that the little group \(\mathrm{H}_k = \mathrm{E}_2\), the Euclidean group in 2 dimensions. Again, it follows that we also have that \(\mathrm{H}_p = \mathrm{E}_2\).

Demanding that \(V_p\), or equivalently \(V_k\), is a finite-dimensional representation of the little group, it follows that massless on-shell physical states transform in irreducible representations labelled by their charge under \(\mathrm{SO}(2) \subset \mathrm{E}_2\), with the non-compact generators of \(\mathrm{E}_2\) having trivial action. We denote this charge, which is known as the helicity, by \(\lambda\). It can then be shown that \(W_\mu\) is proportional to \(P_\mu\) \[W_\mu = \lambda P_\mu ~.\] This shows that \(\lambda\) is indeed a Lorentz-invariant quantity. Although, seeing as \(W_\mu\) is a pseudovector while \(P_\mu\) is a vector, under CPT we have that \(\lambda \to -\lambda\). Therefore, by CPT invariance, the irreducible representations will include both states with helicity \(\lambda\) and \(-\lambda\). Finally, while there is no algebraic constraint on \(\lambda\), in physically realistic setups topological considerations constrain it to be either an integer or half-integer.

Having discussed relevant aspects of the Lorentz and Poincaré algebras, we are now in a position to introduce supersymmetry. We start by introducing supersymmetry algebras, which are extensions of the Poincaré algebra by a number of anticommuting supercharges. These are often known as the super-Poincaré algebras and are examples of Lie superalgebras.

To define a Lie superalgebra we start by introducing a \(\mathbb{Z}_2\)-graded vector space. A \(\mathbb{Z}_2\)-graded vector space is a vector space \(V = V_0 \oplus V_1\) together with a \(\mathbb{Z}_2\) grading. Those vectors \(v \in V_0\) or \(v \in V_1\) are known as homogeneous. The grading \(|v|\) of a homogeneous vector \(v\) is given by \[|v| = \begin{cases} 0 & \text{if $v \in V_0$} ~, \\ 1 & \text{if $v \in V_1$} ~. \end{cases}\] In the mathematics literature vectors \(v \in V_0\) are often called even, while vectors \(v \in V_1\) are called odd. In the physics literature, which we will follow here, the adjectives bosonic and fermionic are normally used instead. The usual properties of vector spaces extend to graded vector spaces, always in a way that respects the \(\mathbb{Z}_2\) grading.

A Lie superalgebra is then a \(\mathbb{Z}_2\)-graded vector space \(\mathfrak{g} = \mathfrak{g}_0 \oplus \mathfrak{g}_1\) together with a bracket \[\phantom{}[~,~\}: \mathfrak{g} \times \mathfrak{g} \to \mathfrak{g} ~,\] such that (\(X_i\in\mathfrak{g}_i\), etc.)

it is consistent with the \(\mathbb{Z}_2\) grading \[\phantom{}[X_i,Y_j\} \subset \mathfrak{g}_{i+j\mod2} ~;\]

it has a graded symmetry property \[\phantom{}[X_i,Y_j\} = -(-1)^{ij} [Y_j,X_i\} ~;\]

the graded Jacobi identity is satisfied \[(-1)^{ik} [X_i,[Y_j,Z_k\}\} + (-1)^{ij} [Y_j,[Z_k,X_i\}\} + (-1)^{jk} [Z_k,[X_i,Y_j\}\} = 0 ~.\]

It follows from these properties that \(\mathfrak{g}_0\) is an ordinary Lie algebra with the bracket given by the ordinary commutator \([\mathfrak{g}_0,\mathfrak{g}_0\} = [\mathfrak{g}_0,\mathfrak{g}_0]\). For super-Poincaré algebras \(\mathfrak{g}_0\) always contains at least the Poincaré algebra. The bracket for two elements of \(\mathfrak{g}_1\) becomes the anticommutator \([\mathfrak{g}_1,\mathfrak{g}_1\} = \{\mathfrak{g}_1,\mathfrak{g}_1\}\) (recall \(\{X,Y\} = XY+YX\)). Indeed, the anticommuting supercharges are elements of \(\mathfrak{g}_1\). Finally, the bracket for an element of \(\mathfrak{g}_0\) and an element of \(\mathfrak{g}_1\) is again the ordinary commutator \([\mathfrak{g}_0,\mathfrak{g}_1\} = [\mathfrak{g}_0,\mathfrak{g}_1]\). Moreover, since \([~,~\}: \mathfrak{g}_0 \times \mathfrak{g}_1 \to \mathfrak{g}_1\) the bracket defines an action of \(\mathfrak{g}_0\) on \(\mathfrak{g}_1\) and the graded Jacobi identity implies that \(\mathfrak{g}_1\) transforms in a representation of \(\mathfrak{g}_1\).

Using commutators (\([X,Y] = XY-YX\)) and anticommutators (\(\{X,Y\} = XY+YX\)), write down the four types of Jacobi identity.

The first type of Jacobi identity involves three even generators \[\phantom{}[X_0,[Y_0,Z_0]] + [Y_0,[Z_0,X_0]] + [Z_0,[X_0,Y_0]] = 0 ~.\] The second type of Jacobi identity involves three even generators \[\phantom{}[X_0,[Y_0,Z_1]] + [Y_0,[Z_1,X_0]] + [Z_1,[X_0,Y_0]] = 0 ~.\] The third type of Jacobi identity involves three even generators \[\phantom{}[X_0,\{Y_1,Z_1\}] + \{Y_1,[Z_1,X_0]\} - \{Z_1,[X_0,Y_1]\} = 0 ~.\] The fourth type of Jacobi identity involves three even generators \[\phantom{}[X_1,\{Y_1,Z_1\}] + [Y_1,\{Z_1,X_1\}] + [Z_1,\{X_1,Y_1\}] = 0 ~.\]

The super-Poincaré algebra is a Lie superalgebra with \[\begin{split} \mathfrak{g}_0 : & \qquad \text{Poincaré algebra ( + R-symmetry + central $\mathrm{U}(1)$s)} ~, \\ \mathfrak{g}_1 : & \qquad \text{Supercharges: $Q^I_\alpha$, $\bar Q_{I\dot\alpha} = (Q^I_\alpha)^\dagger$, $I=1,\dots,\mathcal{N}$} ~. \end{split}\] The integer \(\mathcal{N}\) counts the number of supersymmetries. When \(\mathcal{N} \geq 2\) we say that we have extended supersymmetry. From the algebraic point of view there is no limit of \(\mathcal{N}\); however, we will see that limits on the spin of the particles or fields constrains \(\mathcal{N} \leq 4\) (or \(\mathcal{N} \leq 8\) in theories of supergravity).

In addition to those of the Poincaré algebra \(\eqref{eq:poincare}\), we have the new commutation relations \[\begin{equation} \label{eq:comm}
\phantom{}[P_\mu,Q_\alpha^I] = [P_\mu,\bar Q_{I\dot\alpha}] = 0 ~,
\qquad
[M_{\mu\nu},Q_\alpha^I] = i (\sigma_{\mu\nu})_\alpha{}^\beta Q_\beta^I ~, \qquad
[M_{\mu\nu},\bar Q_I^{\dot\alpha}] = i (\bar\sigma_{\mu\nu})^{\dot\alpha}{}_{\dot\beta} \bar Q_I^{\dot\beta} ~. \end{equation}\] The superalgebra is then (almost) completed with the following anticommutation relations between the supercharges \[\begin{equation} \label{eq:acom}
\{Q_\alpha^I,Q_\beta^J\} = \epsilon_{\alpha\beta} Z^{IJ} ~, \qquad
\{\bar Q_{I\dot\alpha},\bar Q_{J\dot\beta}\} = \epsilon_{\dot\alpha\dot\beta} \bar Z_{IJ} ~, \qquad
\{Q_\alpha^I,\bar Q_{J\dot\beta}\} = 2 (\sigma^\mu)_{\alpha\dot\beta} P_\mu \delta^I_J ~, \end{equation}\] where \(Z^{IJ} = - Z^{JI}\) and \(\bar Z_{IJ} = (Z^{IJ})^\dagger\) are central charges, i.e. they commute with all other generators. Finally, there may be an additional bosonic symmetry known as an R-symmetry acting on the index \(I\). The R-symmetry can at most be \(\mathrm{U}(\mathcal{N})\) with \(Q^I_\alpha\) transforming in the fundamental representation and \(\bar Q_{I\dot\alpha}\) in the antifundamental. When \(Z^{IJ} = 0\) the R-symmetry is an automorphism of the commutation and anticommutation relations \(\eqref{eq:comm}\) and \(\eqref{eq:acom}\). Whether or not the R-symmetry is actually a symmetry depends on the specific theory under consideration, e.g. the form of the interactions and the vanishing or otherwise of the central charges. ^{4}

Here we have simply presented the commutation and anticommutation relations of the super-Poincaré algebra. However, it is possible to argue that its form is uniquely fixed by requiring that ^{5}

it contains the Poincaré algebra as a subalgebra;

it is a graded Lie algebra, or Lie superalgebra, satisfying the graded Jacobi identity;

there are no conserved charges of spin greater than one.

The final requirement is physically motivated by the Coleman-Mandula theorem and its generalisations. A sketch of the argument goes as follows. The Coleman-Mandula theorem states that (under certain assumptions) there are no conserved charges of integer spin greater than one. Now consider the anticommutator of a supercharge of spin \(\frac{n}{2}\), where \(n \in \mathbb{Z}\), with its conjugate. By positivity this anticommutator can only vanish if the supercharge itself vanishes. Therefore, if it is non-vanishing it will give a charge of spin \(n\). Seeing as we already know that \(n\) can be at most one, it follows that only supercharges with spin \(\frac{1}{2}\) are allowed.

Supersymmetric quantum mechanics with two supercharges (sometimes called \(\mathcal{N}=2\) or \(\mathcal{N}=(1,1)\)) has the superalgebra \[\begin{gathered} \{Q,\bar Q\} = 2H ~, \qquad \{Q,Q\} = Z ~, \qquad \{\bar Q,\bar Q\} = \bar Z ~, \\ \phantom{}[R,Q] = -Q ~, \qquad [R,\bar Q] = \bar Q ~, \qquad [R,Z] = -2Z ~, \qquad [R,\bar Z] = 2\bar Z ~, \end{gathered}\] where \(H\) is the Hamiltonian and all other commutators vanish. Check the graded Jacobi identity for this superalgebra.

We start by observing that \(H\) is central, hence all Jacobi identities involving \(H\) are trivially satisfied. It remains to check those Jacobi identities involving one of the seven non-vanishing commutators or anticommutators and one of \(Q\), \(Z\) or \(R\): \[\begin{split} & [Q,\{Q,\bar Q\}] + [Q,\{\bar Q, Q\}] + [\bar Q,\{Q,Q\}] = 4[Q,H] + [\bar Q,Z] = 0 ~, \\ & [Q,\{Q,Q\}] + [Q,\{Q, Q\}] + [Q,\{Q,Q\}] = 3[Q,Z] = 0 ~, %\\ %& [Q,\{\bar Q,\bar Q\}] + [\bar Q,\{\bar Q, Q\}] + [\bar Q,\{Q,\bar Q\}] = [Q,\bar Z] + 4[\bar Q,H] = 0 ~, \\ & \{Q,[R,Q]\} - [R,\{Q,Q\}] - \{Q,[Q,R]\} = -\{Q,Q\} - [R,Z] - \{Q,Q\} = -2Z + 2Z = 0 ~, \\ & \{Q,[R,\bar Q]\} - [R,\{\bar Q,Q\}] - \{\bar Q,[Q,R]\} = \{Q,\bar Q\} - 2[R,H] - \{\bar Q,Q\} = 0 ~, \\ & [Q,[R,Z]] + [R,[Z,Q]] + [Z,[Q,R]] = -2[Q,Z] + [Z,Q] = 0 ~, \\ & [Q,[R,\bar Z]] + [R,[\bar Z,Q]] + [\bar Z,[Q,R]] = 2[Q,\bar Z] + [\bar Z,Q] = 0 ~, \\ & [Z,\{Q,\bar Q\}] + \{Q,[\bar Q,Z]\} - \{\bar Q,[Z,Q]\} = 2[Z,H] = 0 ~, \\ & [Z,\{Q,Q\}] + \{Q,[Q,Z]\} - \{Q,[Z,Q]\} = [Z,Z] = 0 ~, \\ & [Z,\{\bar Q,\bar Q\}] + \{\bar Q,[\bar Q,Z]\} - \{\bar Q,[Z,\bar Q]\} = [Z,\bar Z] = 0 ~, %\\ %& [Z,[R,Q]] + [R,[Q,Z]] + [Q,[Z,R]] = -[Z,Q] + 2[Q,Z] = 0 ~, %\\ %& [Z,[R,\bar Q]] + [R,[\bar Q,Z]] + [\bar Q,[Z,R]] = [Z,\bar Q] + 2[\bar Q,Z] = 0 ~, \\ & [Z,[R,\bar Z]] + [R,[\bar Z,Z]] + [\bar Z,[Z,R]] = 2[Z,\bar Z] + 2[\bar Z,Z] = 0 ~. %\\ %& [R,\{Q,\bar Q\}] + \{Q,[\bar Q,R]\} - \{\bar Q,[R,Q]\} = 2[R,H] - \{Q,\bar Q\} + \{\bar Q,Q\} = 0 ~, %\\ %& [R,\{Q,Q\}] + \{Q,[Q,R]\} - \{Q,[R,Q]\} = [R,Z] + \{Q,Q\} + \{Q,Q\} = -2Z + Z + Z = 0 ~. \end{split}\] The remaining Jacobi identities are either trivially satisfied, e.g. \([X_0,[X_0,Y_i]] + [X_0,[Y_i,X_0]] + [Y_i,[X_0,X_0]] = 0\), or are related to those above by conjugation.

**Basic consequences of the super-Poincaré algebras.**

Now we have introduced the super-Poincaré algebras, let us look at some of their basic consequences.

First, all superpartners (fields or particles related by the action of the supercharges) have the same mass. This is a consequence of the commutation relations \([P_\mu,Q_\alpha^I] = [P_\mu,\bar Q_{I\dot\alpha}] = 0\), which in turn implies that \([P^2,Q_\alpha^I] = [P^2,\bar Q_{I\dot\alpha}] = 0\) and \(P^2\) is still a Casimir invariant. The mass-squared is the eigenvalue of \(P^2\).

Second, the energy of any physical state is non-negative, and it is zero only for a supersymmetric ground state. To see this, let us consider a general state \(|\Phi\rangle\). By the positivity of the Hilbert space we have \[\begin{split} 0 \leq ||Q_\alpha^1|\Phi\rangle||^2 + || \bar Q_{1\dot\alpha}|\Phi\rangle||^2 & = \langle\Phi|\big(\bar Q_{1\dot\alpha}Q_\alpha^1 + Q_\alpha^1\bar Q_{1\dot\alpha}\big) |\Phi\rangle \\ & = \langle\Phi| \{\bar Q_{1\dot\alpha},Q_\alpha^1\} |\Phi\rangle = 2 (\sigma^{\mu})_{\alpha\dot\alpha} \langle\Phi|P_\mu |\Phi\rangle ~. \end{split}\] Now contracting with \(\delta^{\alpha\dot\alpha}\), which amounts to taking the trace of the matrix \(\sigma^\mu\), we find \[0 \leq 2\mathop{\mathrm{tr}}(\sigma^\mu)\langle\Phi|P_\mu |\Phi\rangle =4 \langle\Phi|P_0 |\Phi\rangle ~.\] Therefore, the energy of any physical state is positive. Note that we could have used any supercharge \(Q_\alpha^I\), together with its conjugate \(\bar Q_{I\dot\alpha}\), to derive this result. To show the converse we start by writing \[\langle\Phi|P_0 |\Phi\rangle = \frac{1}{4}\Big(\sum_\alpha ||Q_\alpha^I|\Phi\rangle||^2 + \sum_{\dot\alpha} ||\bar Q_{I\dot\alpha}|\Phi\rangle||^2 \Big) ~.\] Again by the positivity of the Hilbert space, this vanishes if and only if \[Q_\alpha^I|\Phi\rangle = \bar Q_{I\dot\alpha}|\Phi\rangle = 0 \qquad \text{for all $\alpha$, $\dot\alpha$, $I$} ~.\] That is the state \(\Phi\) is supersymmetric. Given that these statements follow only from the supersymmetry algebra they are protected against quantum corrections.

Third, any physical supermultiplet (apart from the vacuum) contains an equal number of on-shell bosonic and fermionic degrees of freedom. Consider the fermion number operator \(F\), defined such that \[(-1)^F |\text{Boson}\rangle = |\text{Boson}\rangle ~, \qquad
(-1)^F |\text{Fermion}\rangle = - |\text{Fermion}\rangle ~.\] Since a supercharge maps bosonic states to fermionic states and vice versa, it follows that \[\begin{equation} \label{eq:idfer}
(-1)^F Q = - Q (-1)^F ~, \end{equation}\] and similarly for the conjugate supercharge. Let us now consider a supermultiplet at fixed momentum \(p_\mu\). The on-shell degrees of freedom should then transform in a finite-dimensional irreducible representation of the subalgebra of the super-Poincaré algebra that leaves this momentum invariant. ^{6} Seeing as \([P_\mu,Q_\alpha^I] = [P_\mu,\bar Q_{I\dot\alpha}] = 0\) this subalgebra includes the supercharges. Taking the trace over the representation we have \[\begin{split}
0 & = \mathop{\mathrm{tr}}\big(-Q_\alpha^I(-1)^F\bar Q_{J\dot\beta} + (-1)^F \bar Q_{J\dot\beta}Q_\alpha^I\big)
\\
& = \mathop{\mathrm{tr}}\big((-1)^F\{Q_\alpha^I,\bar Q_{J\dot\beta}\}\big) = 2 (\sigma^\mu)_{\alpha\dot\beta} \delta_I^J \mathop{\mathrm{tr}}\big((-1)^F P_\mu\big)
\\
& = 2 (\sigma^\mu)_{\alpha\dot\beta} \delta^I_J p_\mu \mathop{\mathrm{tr}}\big((-1)^F\big) ~,
\end{split}\] where the first equality holds by the cyclicity of the trace and to go from the first to second line we use that \((-1)^F\) anticommutes with supercharges \(\eqref{eq:idfer}\). Now choosing any \(p_\mu \neq 0\), which is possible if and only if \(E = p_0 \neq 0\), we find \[\mathop{\mathrm{tr}}\big((-1)^F\big) = \#_{\text{Boson}} - \#_{\text{Fermion}} = 0 ~.\] This holds for any physical supermultiplet that is not the vacuum (\(E\neq0\)).

Supermultiplets are unitary irreducible representations of the super-Poincaré algebra. Since the Poincaré algebra is a subalgebra of the super-Poincaré algebra, they are also (typically reducible) representations of the Poincaré algebra. By definition, physical supermultiplets are on-shell. ^{7} Just as for the unitary irreducible representations of the Poincaré algebra, we will focus our analysis on the two most physically relevant cases: positive-energy massless and massive supermultiplets.

To study the massless supermultiplets we start by fixing the Lorentz frame \[\begin{equation} \label{eq:lorentzframemassless} p_\mu = k_\mu = (E,0,0,E)_\mu \qquad \Rightarrow \qquad \sigma^\mu p_\mu = \begin{pmatrix} 0 & 0 \\ 0 & 2E \end{pmatrix} ~. \end{equation}\] It follows that \[\{Q^I_\alpha,\bar Q_{J\dot\beta}\} = \begin{pmatrix} 0 & 0 \\ 0 & 4E \end{pmatrix}_{\alpha\dot\beta} \delta^I_J \qquad \Rightarrow \qquad \{Q^I_1,\bar Q_{J1}\} = 0 \qquad \text{for all $I$, $J$} ~.\] Therefore, we have that \(||Q_1^I|\Phi\rangle||^2 + ||\bar Q_{I1} |\Phi\rangle||^2 = 0\) for all states \(|\Phi\rangle\) in the supermultiplet. By the positivity of the Hilbert space it follows that \(Q_1^I = \bar Q_{I1} = 0\) on this supermultiplet. In addition, this implies that the central charges \(Z^{IJ}\) and \(\bar Z_{IJ}\) vanish on the supermultiplet.

We have seen that half of the supercharges annihilate the supermultiplet. For the remaining half, we define \[a^I = \frac{1}{2\sqrt{E}} Q_2^I ~, \qquad a_I^\dagger = \frac{1}{2\sqrt{E}} \bar Q_{I2} ~,\] which satisfy the anticommutation relations \[\{a^I,a_J^\dagger\} = \delta^I_J ~, \qquad \{a^I,a^J\} = \{a_I^\dagger,a_J^\dagger\} = 0 ~.\] These are nothing but \(\mathcal{N}\) fermionic annihilation and creation operators. In the Lorentz frame \(\eqref{eq:lorentzframemassless}\) the generator of the \(\mathrm{SO}(2)\) subgroup of the little group is \(J_3 = M_{12}\). The commutation relations \(\eqref{eq:comm}\) imply that \[\phantom{}[M_{12},a^I] = -\frac{1}{2}a^I ~, \qquad [M_{12},a^\dagger_I] = + \frac{1}{2} a^\dagger_I ~,\] hence \(a^I\) lowers the helicity by \(\frac{1}{2}\) while \(a^\dagger_I\) raises it by \(\frac{1}{2}\).

To build supermultiplets we start from a Clifford vacuum \(|\lambda_0\rangle\) with helicity \(\lambda_0\) that is annihilated by all \(a^I\)s \[a^I|\lambda_0\rangle = 0 \qquad \text{for all $I = 1,\dots,\mathcal{N}$} ~,\] and acting on it with the \(\mathcal{N}\) creation operators \(a_I^\dagger\) \[\begin{split} & |\lambda_0\rangle ~, \\ a_I^\dagger & |\lambda_0\rangle = |\lambda_0 + \frac{1}{2}\rangle_I ~, \\ a_I^\dagger a_J^\dagger & |\lambda_0\rangle = |\lambda_0 + 1\rangle_{[IJ]} ~, \\ & \vdots \\ a_1^\dagger a_2^\dagger \dots a_{\mathcal{N}}^\dagger &|\lambda_0\rangle = |\lambda_0 + \frac{\mathcal{N}}{2}\rangle ~. \end{split}\] Since each creation operator can only be applied once, it is easy to see that the total number of states in the supermultiplet is \(2^{\mathcal{N}}\) and that there are \({\mathcal{N} \choose k}\) states of helicity \(\lambda_0+\frac{k}{2}\). Indeed, we have that \(\sum_{k=0}^{\mathcal{N}} {\mathcal{N} \choose k} = (1+1)^{\mathcal{N}} = 2^{\mathcal{N}}\). As a consistency check we can also see that there are an equal number of bosonic and fermionic states in the supermultiplet by computing \(\mathop{\mathrm{tr}}\big((-1)^F\big)\). Doing so, we find \[\mathop{\mathrm{tr}}\big((-1)^F\big) = \sum_{k=0}^{\mathcal{N}} {\mathcal{N} \choose k} (-1)^{2\lambda_0+k} = (-1)^{2\lambda_0} (1-1)^\mathcal{N} = 0 ~.\]

Finally, recalling that under CPT the helicity of a state changes sign, the supermultiplet that we have constructed can only be CPT-invariant if \(\lambda_0 = -\frac{\mathcal{N}}{4}\). In this case we say that the supermultiplet is self-conjugate under CPT. If it is not self-conjugate, then we add the CPT-conjugate to construct a physical supermultiplet with \(2^{\mathcal{N}+1}\) degrees of freedom.

**\(\mathcal{N}=1\) massless supermultiplets.**

The \(\mathcal{N}=1\) massless supermultiplet contains two states, or four when we include the CPT-conjugate. For \(\mathcal{N}=1\) it is straightforward to see that a supermultiplet on its own cannot be self-conjugate. We denote the content of a supermultiplet of helicity \(\lambda_0\) and its CPT-conjugate as follows: \[\Big(\lambda_0,\lambda_0+\frac{1}{2}\Big) + \Big(-\lambda_0,-\lambda_0-\frac{1}{2}\Big) = \Big(-\lambda_0-\frac{1}{2},-\lambda_0,\lambda_0,\lambda_0+\frac{1}{2}\Big) ~.\] Restricting to those relevant for SQFT and supergravity, ^{8} the possible \(\mathcal{N}=1\) massless supermultiplets are:

Chiral multiplet (\(\lambda_0 = 0\)): \(\big(0,\frac{1}{2}\big) + \big(0,-\frac{1}{2}\big) = \big(-\frac{1}{2},0,0,\frac{1}{2}\big)\), containing one Weyl fermion and one complex scalar;

Vector multiplet (\(\lambda_0 = \frac{1}{2}\)): \(\big(\frac{1}{2},1\big)+\big(-\frac{1}{2},-1\big) = \big(-1,-\frac{1}{2},\frac{1}{2},1\big)\), containing one gauge boson and one Weyl fermion;

Gravitino multiplet (\(\lambda_0 = 1\)): \(\big(1,\frac{3}{2}\big) + \big(-1,-\frac{3}{2}\big) = \big(-\frac{3}{2},-1,1,\frac{3}{2}\big)\), containing one gravitino and one gauge boson;

^{9}Graviton multiplet (\(\lambda_0 = \frac{3}{2}\)): \(\big(\frac{3}{2},2\big) + \big(-2,-\frac{3}{2}\big) = \big(-2,-\frac{3}{2},\frac{3}{2},2\big)\), containing one graviton and one gravitino.

**\(\mathcal{N}=2\) massless supermultiplets.**

The \(\mathcal{N}=2\) massless supermultiplet contains four states, or eight when we include the CPT-conjugate. Restricting to those relevant for SQFT, the possible \(\mathcal{N}=2\) massless supermultiplets are:

Hypermultiplet (\(\lambda_0 = -\frac{1}{2}\)): \(\big(2 \times (-\frac{1}{2}), 4\times (0),2\times(\frac{1}{2})\big)\), containing two \(\mathcal{N}=1\) chiral multiplets in complex conjugate representations;

\(\mathcal{N}=2\) vector multiplet (\(\lambda = 0\)): \(\big(-1,2 \times (-\frac{1}{2}), 2\times (0),2\times(\frac{1}{2}),1\big)\), containing one \(\mathcal{N}=1\) vector multiplet and one \(\mathcal{N}=1\) chiral multiplet.

For \(\mathcal{N}=2\) we can also construct a self-conjugate multiplet with \(\lambda_0 = -\frac{1}{2}\), the half-hypermultiplet, which contains the same content as the \(\mathcal{N}=1\) chiral multiplet: \(\big(-\frac{1}{2},2\times (0),\frac{1}{2}\big)\). That the half-hypermultiplet is self-conjugate is not guaranteed and leads to constraints on the representations of the gauge and flavour groups in which it transforms. Often it is more straightforward to think of half-hypermultiplets as constrained hypermultiplets, i.e. hypermultiplets subject to a particular reality condition.

Write down the content of the \(\mathcal{N}=2\) gravitino (\(\lambda_0 = \frac{1}{2}\)) and graviton (\(\lambda_0 = 1\)) multiplets. Decompose these supermultiplets in terms of \(\mathcal{N}=1\) massless supermultiplets.

The content of the gravitino multiplet is \(\big(\frac{1}{2},2\times(1),\frac{3}{2}\big)\) plus its CPT-conjugate \(\big(-\frac{3}{2},2\times(-1),-\frac{1}{2}\big)\). Combined this gives \(\big(-\frac{3}{2},2\times(-1),-\frac{1}{2},\frac{1}{2},2\times(1),\frac{3}{2}\big)\) and contains one \(\mathcal{N}=1\) gravitino multiplet and one \(\mathcal{N}=1\) vector multiplet. The content of the graviton multiplet is \(\big(1,2\times(\frac{3}{2}),2\big)\) plus its CPT-conjugate \(\big(-2,2\times(-\frac{3}{2}),-1\big)\). Combined this gives \(\big(-2,2\times(-\frac{3}{2}),-1,1,2\times(\frac{3}{2}),2\big)\) and contains one \(\mathcal{N}=1\) graviton multiplet and one \(\mathcal{N}=1\) gravitino multiplet.

**\(\mathcal{N}=4\) massless supermultiplets.**

For \(\mathcal{N}=4\) there is only a single supermultiplet relevant for SQFT. This is the vector multiplet (\(\lambda_0 = -1\)): \(\big(-1,4\times(-\frac{1}{2}),6\times(0),4\times(\frac{1}{2}),1)\), which contains one \(\mathcal{N}=2\) vector multiplet and one \(\mathcal{N}=2\) hypermultiplet. It is a self-conjugate supermultiplet. With massless states of helicity \(\pm1\) it is natural to expect vector multiplets to originate from gauge theories; hence all the states should transform in the adjoint representation of the gauge group. Since for \(\mathcal{N}=4\) there is only the vector multiplet and no matter multiplet, there is a unique \(\mathcal{N}=4\) SQFT, up to the choice of gauge group, known as \(\mathcal{N}=4\) super-Yang-Mills.

We have skipped \(\mathcal{N}=3\) because the only supermultiplet relevant for SQFT is the \(\mathcal{N}=3\) vector supermultiplet, which should be considered together with its CPT-conjugate. Combined these have the same content as the \(\mathcal{N}=4\) vector supermultiplet. In fact, it transpires that any \(\mathcal{N}=3\) SQFT that has a Lagrangian is actually \(\mathcal{N}=4\) super-Yang-Mills. There are \(\mathcal{N}=3\) SCFTs that do not have \(\mathcal{N}=4\) supersymmetry; however, these do not have a Lagrangian description.

Explicitly construct the most general massless supermultiplet in a SQFT with \(\mathcal{N}=3\) supersymmetry. What is the content of this supermultiplet? Show that it coincides with the content of the \(\mathcal{N} =4\) vector multiplet.

Starting with a Clifford vacuum of helicity \(\lambda_0\) we construct the supermultiplet as follows: \[|\lambda_0\rangle ~, \qquad a_I^\dagger |\lambda_0\rangle ~, \qquad a_{[J}^\dagger a_{K]}^\dagger |\lambda_0\rangle ~, \qquad a_1^\dagger a_2^\dagger a_3^\dagger |\lambda_0\rangle ~, \qquad I=1,2,3~,\] which contains \(1+3+3+1=8\) states. Since each creation operator raises the helicity of the state by \(\frac{1}{2}\), the content of this supermultiplet is \(\big(\lambda_0,3\times(\lambda_0+\frac{1}{2}),3\times(\lambda_0+1),\lambda_0+\frac{3}{2}\big)\). Recalling that under CPT the helicity of a state changes sign, this supermultiplet cannot be self-conjugate with integer or half-integer \(\lambda_0\). Therefore, we also need to include its CPT-conjugate, which has content \(\big(-\lambda_0-\frac{3}{2},3\times(-\lambda_0-1),3\times(-\lambda_0-\frac{1}{2}),-\lambda_0\big)\). In SQFTs supermultiplets cannot have states with \(|\lambda| > 1\). This immediately implies that \(\lambda_0 = -\frac{1}{2}\), and the total content of the \(\mathcal{N}=3\) vector multiplet is \(\big(-1,4\times(-\frac{1}{2}),6\times (0),4\times(\frac{1}{2}),1\big)\), hence it contains one massless vector, four massless Weyl fermions and six massless scalars. This is the same content as the \(\mathcal{N}=4\) vector multiplet.

If we take \(\mathcal{N}>4\) then the supermultiplet necessarily contains states with \(|\lambda|>1\). Therefore, we no longer have a SQFT, but rather a theory of supergravity. For \(\mathcal{N}>8\) the supermultiplet necessarily contains states with \(|\lambda|>2\). Consequently, \(\mathcal{N}=8\) is usually considered to be the maximum number of supersymmetries since it is believed that no massless particles or fields with \(|\lambda|>2\) exist.

\(\mathcal{N}=8\) is special because there is a single massless supermultiplet for which all states have helicity \(|\lambda| \leq 2\). This is the \(\mathcal{N}=8\) graviton multiplet. Find the content of the \(\mathcal{N}=8\) graviton multiplet. Decompose the \(\mathcal{N}=8\) graviton multiplet in terms of \(\mathcal{N}=1\) massless supermultiplets.

Starting with a Clifford vacuum of helicity \(\lambda_0\) the supermultiplet has content \(\big(\lambda_0,8\times(\lambda_0+\frac{1}{2}),28\times(\lambda_0+1),56\times(\lambda_0+\frac{3}{2}),70\times(\lambda_0+2),56\times(\lambda_0+\frac{5}{2}),28\times(\lambda_0+3),8\times(\lambda_0+\frac{7}{2}),\lambda_0+4\big)\). Requiring that there are no states with \(|\lambda|>2\) implies that \(\lambda_0 = -2\), which also means that we have a self-conjugate supermultiplet. Therefore, the content of the \(\mathcal{N}=8\) graviton multiplet is \(\big(-2,8\times(-\frac{3}{2}),28\times(-1),56\times(-\frac{1}{2}),70\times(0),56\times(\frac{1}{2}),28\times(1),8\times(\frac{3}{2}),2\big)\), hence it contains 1 graviton, 8 gravitinos, 28 vectors, 56 Weyl fermions and 70 scalars. In terms of \(\mathcal{N}=1\) massless supermultiplets this is 1 graviton multiplet, 7 (\(= 8-1\)) gravitino multiplets, 21 (\(=28-7\)) vector multiplets and 35 (\(=56-21\)) chiral multiplets. As a consistency check, the 35 chiral multiplets indeed contain 70 scalars.

To study the massive supermultiplets we start by fixing the Lorentz frame \[\begin{equation} \label{eq:lorentzframemassive} p_\mu = k_\mu = (m,0,0,0)_\mu \qquad \Rightarrow \qquad \sigma^\mu p_\mu = \begin{pmatrix} m & 0 \\ 0 & m \end{pmatrix} ~. \end{equation}\] It follows that the anticommutation relations of the supercharges are given by \[\begin{equation} \label{eq:acommassive} \{Q^I_\alpha,\bar Q_{J\dot\beta}\} = 2m \delta_{\alpha\dot\beta}\delta^I_J ~, \qquad \{Q^I_\alpha,Q^J_\beta\} = \epsilon_{\alpha\beta}Z^{IJ} ~, \qquad \{\bar Q_{I\dot\alpha},\bar Q_{J\dot\beta}\} = \epsilon_{\dot\alpha\dot\beta} \bar Z_{IJ} ~. \end{equation}\]

**\(\mathcal{N}=1\) massive supermultiplets.**

For \(\mathcal{N}=1\) there are no central charges. Therefore, introducing \[a_\alpha = \frac{1}{\sqrt{2m}} Q_\alpha ~, \qquad a_{\dot\alpha}^\dagger = \frac{1}{\sqrt{2m}} \bar Q_{\dot\alpha} ~,\] we see that we have twice as many fermionic oscillators compared to the corresponding massless case \[\{a_\alpha,a_{\dot\beta}^\dagger\} = \delta_{\alpha\dot\beta} ~, \qquad \{a_\alpha,a_\beta\} = 0 ~, \qquad \{a_{\dot\alpha}^\dagger,a_{\dot\beta}^\dagger\} = 0 ~.\] In the Lorentz frame \(\eqref{eq:lorentzframemassive}\) the generators of the \(\mathrm{SO}(3)\) little group are \(J_i = \frac{1}{2}\epsilon_{ijk}M_{jk}\). Working in an eigenbasis of \(J_3 = M_{12}\), we call the eigenvalue of \(J_3\) the \(z\)-component of the spin. The commutation relations \(\eqref{eq:comm}\) then imply that \[\phantom{}[M_{12},a_1] = \frac{1}{2} a_1 ~, \qquad [M_{12},a_2] = -\frac{1}{2} a_2 ~, \qquad [M_{12},a_1^\dagger] = -\frac{1}{2} a_1^\dagger ~, \qquad [M_{12},a_2^\dagger] = \frac{1}{2} a_2^\dagger ~,\] hence \(a_1\) and \(a_2^\dagger\) raise the \(z\)-component of the spin by \(\frac{1}{2}\), while \(a_2\) and \(a_1^\dagger\) lower it by \(\frac{1}{2}\).

To build supermultiplets we start from a Clifford vacuum \(|m,s;s_z\rangle\) of mass \(m\) and spin \(s\) that is annihilated by \(a_1\) and \(a_2\). Such a Clifford vacuum is comprised of \(2s+1\) states labelled by \(s_z = \{-s,-s+1,\dots,s-1,s\}\). There are two supermultiplets that are relevant for SQFT. ^{10} These are the \(\mathcal{N}=1\) massive chiral multiplet (\(s=0\)):

containing one massive Majorana fermion and one massive complex scalar, and the \(\mathcal{N}=1\) massive vector multiplet (\(s=\frac{1}{2}\)):

containing one massive vector, one massive Dirac fermion and one massive real scalar. The content of the \(\mathcal{N}=1\) massive vector multiplet is consistent with a super-Higgs mechanism: a massless vector multiplet combines with a massless chiral multiplet to become a massive vector multiplet.

**\(\mathcal{N} \geq 2\) massive supermultiplets.**

For \(\mathcal{N} \geq 2\) massive supermultiplets we can have non-vanishing central charges. By a \(U(\mathcal{N})\) (R-symmetry) rotation, we can always choose the central charge matrix \(Z^{IJ}\), which we recall is antisymmetric, to take its real normal form \[\begin{equation} \label{eq:ccmat}
Z^{IJ} = \left(\begin{smallmatrix}
0 & z_1 & & & & &
\\ -z_1 & 0 & & & & &
\\ & & 0 & z_2 & & &
\\ & & -z_2 & 0 & & &
\\ & & & & \ddots & &
\\ & & & & & 0 & z_{\frac{\mathcal{N}}{2}}
\\ & & & & & -z_{\frac{\mathcal{N}}{2}} & 0
\end{smallmatrix}\right)^{IJ} ~,
\qquad
Z^{IJ} = \left( \begin{smallmatrix}
0 & z_1 & & & & & &
\\ -z_1 & 0 & & & & & &
\\ & & 0 & z_2 & & & &
\\ & & -z_2 & 0 & & & &
\\ & & & & \ddots & & &
\\ & & & & & 0 & z_{\frac{\mathcal{N}-1}{2}} &
\\ & & & & & -z_{\frac{\mathcal{N}-1}{2}} & 0 &
\\ & & & & & & & 0
\end{smallmatrix}\right)^{IJ} ~, \end{equation}\] where \(z_r\) are real and non-negative, for even and odd \(\mathcal{N}\) respectively. Working with a pair of supercharges for each \(z_r\) (if \(\mathcal{N}\) is odd, the last supercharge is treated as for \(\mathcal{N}=1\)), we define \[a_\alpha^r = \frac{1}{\sqrt{2}}(Q_\alpha^{2r-1} + \epsilon_{\alpha}{}^{\dot\beta} \bar Q_{2r\dot\beta}) ~,\qquad
b_\alpha^r = \frac{1}{\sqrt{2}}(Q_\alpha^{2r-1} - \epsilon_{\alpha}{}^{\dot\beta} \bar Q_{2r\dot\beta}) ~, \qquad
r = 1,\dots,\Big\lfloor\frac{\mathcal{N}}{2}\Big\rfloor ~,\] where \(\epsilon_{1}{}^2 = - \epsilon_{2}{}^1 = - 1\) and \(\epsilon_1{}^1 = \epsilon_2{}^2 = 0\), together with their conjugates. Substituting into \(\eqref{eq:acommassive}\), these satisfy the anticommutation relations \[\begin{equation} \label{eq:acommassive2}
\{a^r_\alpha,(a_\beta^s)^\dagger\} = (2m+z_r) \delta^{rs}\delta_{\alpha\beta} ~, \qquad
\{b^r_\alpha,(b_\beta^s)^\dagger\} = (2m-z_r) \delta^{rs}\delta_{\alpha\beta} ~, \end{equation}\] with all other anticommutators vanishing. We have a pair of fermionic annihilation and creation operators for each \(\mathcal{N}\). ^{11} As for \(\mathcal{N}=1\), \(a^r_1\), \((a^r_2)^\dagger\), \(b^r_1\) and \((b^r_2)^\dagger\) raise the \(z\)-component of the spin by \(\frac{1}{2}\), while \(a^r_2\), \((a^r_1)^\dagger\), \(b^r_2\) and \((b^r_1)^\dagger\) lower it by \(\frac{1}{2}\).

Starting from the super-Poincaré algebra with \(\mathcal{N}=2\), fixing the Lorentz frame \(p_\mu = k_\mu = (m,0,0,0)_\mu\) and setting \(Z^{IJ} = \big(\begin{smallmatrix} 0 & z \\ -z & 0 \end{smallmatrix}\big)^{IJ}\), \(z\in\mathbb{R}\) and \(z\geq0\), check that the operators \[a_\alpha = \frac{1}{\sqrt{2}}\big(Q_\alpha^{1} + \epsilon_{\alpha}{}^{\beta} \bar Q_{2\beta}\big) ~,\qquad b_\alpha = \frac{1}{\sqrt{2}}\big(Q_\alpha^{1} - \epsilon_{\alpha}{}^{\beta} \bar Q_{2\beta}\big) ~, \qquad\] satisfy the anticommutation relations \[\{a_\alpha,a_\beta^\dagger\} = (2m+z)\delta_{\alpha\beta} ~, \qquad \{b_\alpha,b_\beta^\dagger\} = (2m-z)\delta_{\alpha\beta} ~,\] with all other anticommutators vanishing. Note that \(\epsilon_{1}{}^2 = - \epsilon_{2}{}^1 = - 1\), \(\epsilon_1{}^1 = \epsilon_2{}^2 = 0\) and we have dropped the distinction between dotted and undotted indices. Working in an eigenbasis of \(J_3 = M_{12}\), determine the effect of the fermionic annihilation and creation operators on the \(z\)-component of the spin (the eigenvalue of \(J_3\)).

Since \(\{Q_\alpha^1,Q_\beta^1\} = \{\bar Q_{2\alpha},\bar Q_{2\beta}\} = 0\) and \(\{Q_\alpha^1,\bar Q_{2\beta}\} = 0\) we have \(\{a_\alpha,a_\beta\} = \{b_\alpha,b_\beta\} = \{a_\alpha,b_\beta\} = 0\). By conjugation this also implies that \(\{a_\alpha^\dagger,a_\beta^\dagger\} = \{b_\alpha^\dagger,b_\beta^\dagger\} = \{a_\alpha^\dagger,b_\beta^\dagger\} = 0\). For the remaining anticommutators we observe that \((\sigma^\mu p_\mu)_{\alpha\beta} = m \delta_{\alpha\beta}\). We then have \[\begin{split} \{a_\alpha,a_\beta^\dagger\} & = \frac{1}{2} \big(\{Q^1_\alpha,\bar Q_{1\beta}\} + \epsilon_{\beta}{}^{\gamma}\{Q^1_\alpha,Q^2_\gamma\} + \epsilon_{\alpha}{}^{\gamma}\{\bar Q_{2\gamma},\bar Q_{1\beta}\} + \epsilon_{\alpha}{}^{\gamma}\epsilon_{\beta}{}^{\delta}\{\bar Q_{2\gamma},Q^2_\delta\} \\ & = \frac{1}{2} \big(2m\delta_{\alpha\beta} + z\epsilon_{\beta}{}^{\gamma}\epsilon_{\alpha\gamma} - z \epsilon_{\alpha}{}^{\gamma}\epsilon_{\gamma\beta} + 2m\epsilon_{\alpha}{}^{\gamma}\epsilon_{\beta}{}^{\delta}\delta_{\gamma\delta}\big) = (2m + z) \delta_{\alpha\beta} ~, \\ \{b_\alpha,b_\beta^\dagger\} & = \frac{1}{2} \big(\{Q^1_\alpha,\bar Q_{1\beta}\} - \epsilon_{\beta}{}^{\gamma}\{Q^1_\alpha,Q^2_\gamma\} - \epsilon_{\alpha}{}^{\gamma}\{\bar Q_{2\gamma},\bar Q_{1\beta}\} + \epsilon_{\alpha}{}^{\gamma}\epsilon_{\beta}{}^{\delta}\{\bar Q_{2\gamma},Q^2_\delta\} \\ & = \frac{1}{2} \big(2m\delta_{\alpha\beta} - z\epsilon_{\beta}{}^{\gamma}\epsilon_{\alpha\gamma} + z \epsilon_{\alpha}{}^{\gamma}\epsilon_{\gamma\beta} + 2m\epsilon_{\alpha}{}^{\gamma}\epsilon_{\beta}{}^{\delta}\delta_{\gamma\delta}\big) = (2m - z) \delta_{\alpha\beta} ~, \\ \{a_\alpha,b_\beta^\dagger\} & = \frac{1}{2} \big(\{Q^1_\alpha,\bar Q_{1\beta}\} - \epsilon_{\beta}{}^{\gamma}\{Q^1_\alpha,Q^2_\gamma\} + \epsilon_{\alpha}{}^{\gamma}\{\bar Q_{2\gamma},\bar Q_{1\beta}\} - \epsilon_{\alpha}{}^{\gamma}\epsilon_{\beta}{}^{\delta}\{\bar Q_{2\gamma},Q^2_\delta\} \\ & = \frac{1}{2} \big(2m\delta_{\alpha\beta} - z\epsilon_{\beta}{}^{\gamma}\epsilon_{\alpha\gamma} - z \epsilon_{\alpha}{}^{\gamma}\epsilon_{\gamma\beta} - 2m\epsilon_{\alpha}{}^{\gamma}\epsilon_{\beta}{}^{\delta}\delta_{\gamma\delta}\big) = 0 ~. \end{split}\] To determine the effect of the fermionic annihilation and creation operators on the \(z\)-component of the spin we compute their commutators with \(M_{12}\). We have \[\begin{aligned} \phantom{}[M_{12},Q_1^I] & = i(\sigma_{12})_1{}^1 Q_1^I = \frac{1}{2} Q_1^I ~, \qquad & \phantom{}[M_{12},Q_2^I] & = i(\sigma_{12})_2{}^2 Q_2^I = -\frac{1}{2} Q_2^I ~, \\ \phantom{}[M_{12},\bar Q_{I1}] & = i\epsilon_{12} (\bar\sigma_{12})^2{}_2 \epsilon^{21}\bar Q_{I1} = - \frac{1}{2} \bar Q_{I1} ~, \qquad & \phantom{}[M_{12},\bar Q_{I2}] & = i\epsilon_{21} (\bar\sigma_{12})^1{}_1 \epsilon^{12}\bar Q_{I2} = \frac{1}{2} \bar Q_{I2} ~. \end{aligned}\] Therefore, \(a_1\), \(a_2^\dagger\), \(b_1\) and \(b_2^\dagger\), which are linear combinations of \(Q^I_1\) and \(\bar Q_{I2}\), raise the \(z\)-component of the spin by \(\frac{1}{2}\), while \(a_2\), \(a_1^\dagger\), \(b_2\) and \(b_1^\dagger\), which are linear combinations of \(Q^I_2\) and \(\bar Q_{I1}\), lower the \(z\)-component of the spin by \(\frac{1}{2}\).

The anticommutation relations \(\eqref{eq:acommassive2}\), together with unitarity, or positivity of the Hilbert space, imply the so-called BPS bound ^{12} \[2m \geq |z_r| \qquad \text{for all $r=1,\dots,\Big\lfloor\frac{\mathcal{N}}{2}\Big\rfloor$} ~.\] If this bound is not satisfied, e.g. \(2m<z_1\), then it is straightforward to show that \(||(b_1^1)^\dagger|0\rangle||^2 = \langle 0| b_1^1 (b_1^1)^\dagger |0\rangle = (2m-z_1) |||0\rangle||^2\), which is negative assuming that the Clifford vacuum has positive norm. If the BPS bound is saturated for some \(r\), then the associated supercharges, either \(a^r_\alpha\) and \((a^r_\alpha)^\dagger\) or \(b^r_\alpha\) and \((b^r_\alpha)^\dagger\), annihilate the supermultiplet, which is shorter as a result. Note that, in the massless case the BPS bound implies \(z_r = 0\) for all \(r\). Therefore, as we have already seen, the central charges vanish on massless supermultiplets. Furthermore, the BPS bound is fully saturated, which is consistent with the fact that half the supercharges annihilate massless supermultiplets.

Let us say that \(k\), where \(0\leq k \leq \lfloor\frac{\mathcal{N}}{2}\rfloor\), of the central charges saturate the bound, and, for the moment, ignore that the Clifford vacuum has spin and is degenerate. If \(k=0\) then the supermultiplet has \(2^{2\mathcal{N}}\) degrees of freedom and we say that we have a long multiplet. If \(k=\frac{\mathcal{N}}{2}\) then the supermultiplet has \(2^{\mathcal{N}}\) degrees of freedom and the multiplet is ultrashort, or \(\frac{1}{2}\)-BPS. For general \(k\), we have a short, or \(\frac{k}{\mathcal{N}}\)-BPS, multiplet with \(2^{2(\mathcal{N}-k)}\) degrees of freedom. The fraction \(\frac{k}{\mathcal{N}}\), where \(0 \leq \frac{k}{\mathcal{N}} \leq \frac{1}{2}\), is the fraction of supercharges that annihilate the supermultiplet. Finally, the count of degrees of freedom coming from the fermionic annihilation and creation operators should be multiplied by the (\(2s+1\)) degeneracy of the Clifford vacuum of spin \(s\).

**Some properties of BPS multiplets.**

The relation \(2m = |z_r|\) is protected against, i.e. not affected by, continuous deformations, e.g. couplings, moduli and quantum corrections in general, because the number of degrees of freedom cannot change. \(m\) and \(z_r\) can change, but only such that \(2m = |z_r|\) still holds. It is possible for two short multiplets form a bound state that is a longer multiplet.

A BPS multiplet can only decay into BPS decay products whose central charges are aligned. As an example, let us take \(\mathcal{N}=2\) and consider the decay of one BPS multiplet, with central charge \(z_{A} \geq 0\) and mass \(m_A\), into two decay products, with central charges \(z_{B}\) and \(z_{C}\) and masses \(m_B\) and \(m_C\). We do not assume that \(z_B\) and \(z_C\) are real and non-negative since we have already used the \(\mathrm{U}(\mathcal{N}=2)\) rotation to choose \(z_A \geq 0\). Conservation of central charge tells us that \(z_{A} = z_{B} + z_{C}\); hence, \(2m_A = |z_{A}| = |z_{B}+z_{C}| \leq |z_{B}| + |z_{C}| \leq 2(m_B+m_C)\), where we have used both the triangle inequality and that the decay products should obey (but not necessarily saturate) the BPS bound. However, on general kinematical grounds, the mass of the initial state in a decay process should be greater than or equal than the sum of the masses of the decay products, i.e. \(m_A \geq m_B + m_C\), which can be most easily seen by going to the rest frame of the decaying particle. Therefore, it follows that \(m_A = m_B + m_C\), hence \(|z_A| = |z_{B}+z_{C}| = |z_{B}| + |z_{C}| = 2(m_B + m_C)\). From this we can conclude that the decay products are themselves BPS multiplets (\(2m_B = |z_B|\) and \(2m_C = |z_C|\)) with aligned central charges (\(z_B\geq 0\) and \(z_C\geq 0\)).

**\(\mathcal{N} = 2\) massive supermultiplets.**

Restricting to those that are relevant for SQFT, the only long \(\mathcal{N}=2\) massive supermultiplet is the vector multiplet (\(s=0\)):

containing one massive vector, two massive Dirac fermions and five massive real scalars. The content of the long \(\mathcal{N}=2\) massive vector multiplet is again consistent with a super-Higgs mechanism: a massless vector multiplet combines with a massless hypermultiplet to become a massive vector multiplet.

For \(\mathcal{N}=2\) there is only a single BPS bound and therefore all short multiplets are \(\frac{1}{2}\)-BPS. Again restricting to those that are relevant for SQFT, the possible short \(\mathcal{N}=2\) massive supermultiplets are:

\(\mathcal{N}=2\) massive hypermultiplet (\(s=0\)):

containing one massive Dirac fermion and two massive complex scalars. The massive hypermultiplet contains a short \(\mathcal{N}=2\) massive multiplet together with its CPT-conjugate. It is possible to construct a self-conjugate multiplet, the massive half-hypermultiplet, which contains one massive complex scalar and one massive Majorana fermion. However, just as in the massless case, this leads to constraints on the representations of the gauge and flavour groups in which it transforms. As \(z \to 0\) the massive hypermultiplet and half-hypermultiplet become their massless counterparts.

Short \(\mathcal{N}=2\) massive vector multiplet (\(s=\frac{1}{2}\)):

containing one massive vector, one massive Dirac fermion and one massive real scalar. This is the same degrees of freedom as a massive \(\mathcal{N}=1\) vector multiplet. The content is again consistent with a super-Higgs mechanism where the massless vector multiplet becomes the short massive vector multiplet.

^{13}As before, as \(z\to0\) the short massive vector multiplet becomes its massless counterpart.

Construct the physical states of the long and short \(\mathcal{N}=2\) massive vector multiplets in terms of oscillators.

For \(\mathcal{N}=2\) we have four fermionic creation operators (\(a_1^\dagger\), \(b_1^\dagger\), \(a_2^\dagger\) and \(b_2^\dagger\)) for the long multiplet and two (\(a_1^\dagger\) and \(a_2^\dagger\)) for the short multiplet. \(a_1^\dagger\) and \(b_1^\dagger\) lower the \(z\)-component of the spin by \(\frac{1}{2}\), while \(a_2^\dagger\) and \(b_2^\dagger\) raise it by \(\frac{1}{2}\).

The physical states of the long \(\mathcal{N}=2\) massive vector multiplet are

The content of the supermultiplet is \(\big(-1,4\times(-\frac{1}{2}),6\times (0),4\times(\frac{1}{2}),1\big)\), hence it contains one massive vector, two massive Dirac fermions and five massive real scalars.

The physical states of the short \(\mathcal{N}=2\) massive vector multiplet are

The content of the supermultiplet is \(\big(-1,2\times(-\frac{1}{2}), 2\times (0), 2\times(\frac{1}{2}),1\big)\), hence it contains one massive vector, one massive Dirac fermion and one massive real scalar.

**\(\mathcal{N}=4\) massive supermultiplets.**

The only \(\mathcal{N}=4\) massive supermultiplet that is relevant for SQFT, is the ultrashort \(\mathcal{N}=4\) massive vector multiplet (\(s=0\)):

containing one massive vector, two massive Dirac fermions and five massive real scalars. The content is again consistent with a super-Higgs mechanism where the massless vector multiplet becomes the ultrashort massive vector multiplet. ^{14} As before, as \(z\to0\) the ultrashort massive vector multiplet becomes its massless counterpart.

Now that we have an understanding of on-shell supermultiplets, we would like to construct the supersymmetric actions that encode their dynamics. Just as for ordinary QFT, to do this we need to introduce off-shell (and potentially gauge) degrees of freedom. Consider a massless vector in 4 dimensions that has two on-shell degrees of freedom with helicities \(\pm1\). An action for these degrees of freedom is given by \(\mathrm{U}(1)\) Yang-Mills theory with the gauge field \(A_\mu\), a Lorentz vector with four components, or off-shell degrees of freedom. One of these is non-dynamical by the \(\mathrm{U}(1)\) gauge symmetry and another by the equations of motion. Another example is a massless Weyl fermion \(\psi_\alpha\) in 4 dimensions, which has two complex, or four real components. However, these obey coupled first-order equation of motions and, as a result, there are only two on-shell degrees of freedom, with helicities \(\pm\frac{1}{2}\).

To construct supersymmetric actions, we would like to have an off-shell linear realisation of supersymmetry that acts on fields. From the content of the on-shell supermultiplets, it is clear that to do this we will need to introduce additional off-shell degrees of freedom to those that we are familiar with from QFT. Consider the \(\mathcal{N}=1\) massless chiral multiplet, which contains two bosonic and two fermionic on-shell degrees of freedom, corresponding to a complex scalar and Weyl fermion. From ordinary QFT, we know that off-shell a complex scalar still has two degrees of freedom, yet a Weyl fermion has four. Therefore, to have an off-shell linear realisation of supersymmetry, we need to introduce two more off-shell bosonic degrees of freedom, in order to match the number of fermionic degrees of freedom. These new degrees of freedom should be non-dynamical and have no kinetic term in the action. We can always eliminate these degrees of freedom, but then supersymmetry will only hold on-shell, i.e. up to the equations of motion.

To proceed we can follow one of two approaches. The first approach is to use components. In terms of components the supercharges are realised as transformations that send bosonic fields to fermionic fields and vice versa – the bosonic and fermionic fields are the components. However, supersymmetry is not manifest, i.e. it cannot be seen by just looking at the action. The second approach is to use superfields. This requires us to introduce superspace, on which the supercharges act as differential operators. Superfields are fields on superspace, which package all the components in one object. The advantages are that supersymmetry is manifest and writing down supersymmetric actions is straightforward. For \(\mathcal{N}=1\) in 4 dimensions using superfields is clearly the more natural approach. However, this approach becomes more and more complicated as \(\mathcal{N}\) increases, and may not always be possible. ^{15}

From this point on we will restrict our discussion to \(\mathcal{N}=1\) supersymmetry in 4 dimensions. We start by introducing \(\mathcal{N}=1\) superspace in 4 dimensions.

Superspace is an extension of ordinary space-time, with coordinates \(x^\mu\), by fermionic coordinates \(\theta_\alpha\) and \(\bar \theta_{\dot \alpha}\), which are anticommuting. Just as \(P_\mu\) generates translations in \(x^\mu\), \(Q_\alpha\) and \(\bar Q_{\dot\alpha}\) generate translations in \(\theta_\alpha\) and \(\bar \theta_{\dot\alpha}\). This should be consistent with the anticommutation relations \[\{Q_\alpha,\bar Q_{\dot\alpha}\} = 2 (\sigma^{\mu})_{\alpha\dot\alpha} P_\mu \qquad \Rightarrow \qquad [\varepsilon_1 Q,\bar \varepsilon_2 \bar Q] = 2(\varepsilon_1\sigma^\mu\bar\varepsilon_2) P_\mu ~,\] where \(\varepsilon_1\) and \(\bar \varepsilon_2\) are constant anticommuting spinors. As a result, \(Q_\alpha\) and \(\bar Q_{\dot\alpha}\) will not only generate translations in \(\theta_\alpha\) and \(\bar \theta_{\dot\alpha}\), but will also act non-trivially on \(x^\mu\).

Let us recall how \(P_\mu\) generates translations in \(x^\mu\), \(x^\mu \to x^\mu + a^\mu\), when acting on a field \(\phi(x)\). Demanding that \[\phi(x+a) = e^{i a^\mu P_\mu} \phi(x) e^{-i a^\mu P_\mu} ~,\] i.e. the field at point \(x^\mu + a^\mu\) is related to the field at point \(x^\mu\) by left conjugation by \(e^{i a^\mu P_\mu}\), we can write both sides as a formal power series in \(a^\mu\) \[\sum_{n=0}^\infty \frac{1}{n!} (a^\mu\partial_\mu)^n \phi(x) = \sum_{n=0}^\infty \frac{1}{n!} (ia^\mu \mathop{\mathrm{ad}}_{P_\mu})^n \phi(x) ~,\] where \(\mathop{\mathrm{ad}}_{P_\mu}\phi(x) = [P_\mu, \phi(x)]\) and \(\partial_\mu = \frac{\partial}{\partial x^\mu}\). Therefore, we immediately see that \(P_\mu\) is realised as a differential operator when it acts on fields ^{16} \[P_\mu = -i \partial_\mu ~.\]

We now perform a similar analysis for the supersymmetry transformations \[\begin{gathered} \theta_\alpha \to \theta_\alpha + \varepsilon_\alpha ~, \qquad \bar\theta_{\dot\alpha} \to \bar\theta_{\dot\alpha} + \bar\varepsilon_{\dot\alpha} ~, \\ x^\mu \to x^\mu + i (\theta\sigma^\mu\bar\varepsilon - \varepsilon \sigma^\mu \bar \theta) ~, \end{gathered}\] where \(\varepsilon_\alpha\) and \(\bar\varepsilon_{\dot\alpha}\) are constant anticommuting spinors parametrising the transformations. The precise form of the transformation of \(x^\mu\) is fixed by the requirement that the supercharges satisfy the anticommutation relations of the superalgebra. A superfield is a field that depends not only on \(x^\mu\), but also on \(\theta_\alpha\) and \(\bar\theta_{\dot\alpha}\) \[Y(x,\theta,\bar\theta) ~.\] Demanding that \[Y(x+ i(\theta\sigma\bar\varepsilon - \varepsilon \sigma \bar \theta),\theta+\varepsilon,\bar\theta+\bar\varepsilon) = e^{i(\varepsilon Q + \bar \varepsilon \bar Q)} Y(x,\theta,\bar\theta) e^{-i(\varepsilon Q + \bar \varepsilon \bar Q)} ~.\] and expanding both sides to leading order in \(\varepsilon\) and \(\bar\varepsilon\) we find \[\begin{split} \big(1+i(\theta\sigma^\mu\bar\varepsilon - \varepsilon \sigma^\mu \bar \theta) \partial_\mu + \varepsilon^\alpha\partial_\alpha + \bar\varepsilon^{\dot\alpha} \bar \partial_{\dot\alpha} + \dots\big) Y(x,\theta,\bar\theta) =\big(1+i\mathop{\mathrm{ad}}_{\varepsilon^\alpha Q_\alpha + \bar \varepsilon_{\dot\alpha} \bar Q^{\dot\alpha}} + \dots\big)Y(x,\theta,\bar\theta) ~ , \end{split}\] where \[\partial_\mu = \frac{\partial}{\partial x^\mu} ~, \qquad \partial_\alpha = \frac{\partial}{\partial \theta^\alpha} ~, \qquad \partial_{\dot\alpha} = \frac{\partial}{\partial \bar\theta^{\dot\alpha}} ~.\] Using Lorentz covariance, we can immediately read off the realisation of \(Q_\alpha\) and \(\bar Q_{\dot \alpha}\) in terms of differential operators \[\begin{equation} \label{eq:qdiffop} Q_\alpha = - i \big(\partial_\alpha -i(\sigma^\mu\bar\theta)_\alpha\partial_\mu\big) ~, \qquad \bar Q_{\dot\alpha} = i \big(\bar\partial_{\dot\alpha} - i (\theta \sigma^\mu)_{\dot\alpha}\partial_\mu\big) ~. \end{equation}\] The differential operators \(\partial_\mu\), \(\partial_\alpha\) and \(\bar\partial_{\dot\alpha}\) satisfy the following relations \[\begin{split} [\partial_\mu,x^\nu] & = \delta_\mu^\nu ~, \qquad \{\partial_\alpha,\theta^\beta\} = \delta_{\alpha}^{\beta} ~, \qquad \{\bar\partial_{\dot\alpha},\bar\theta^{\dot\beta}\} = \delta_{\dot\alpha}^{\dot\beta} ~, \\ [\partial_\alpha,x^\mu] & = [\bar\partial_{\dot\alpha},x^\mu] = [\partial_\mu,\theta^\alpha] = [\partial_\mu,\bar\theta^{\dot\alpha}] = \{\partial_\alpha,\bar\theta^{\dot\beta}\} = \{\bar\partial_{\dot\alpha},\theta^{\beta}\} = 0 ~, \\ [\partial_\alpha,\partial_\mu] & = [\bar\partial_{\dot\alpha},\partial_\mu] = [\partial_\mu,\partial_\nu] = \{\partial_\alpha,\partial_\beta\} = \{\bar\partial_{\dot\alpha},\bar\partial_{\dot\beta}\} = \{\partial_\alpha,\bar\partial_{\dot\beta}\} = 0 ~, \end{split}\] where \(\partial_\alpha\) and \(\bar\partial_{\dot\alpha}\) act on everything to their right. We also have the conjugation relations \((\partial_\alpha)^\dagger = \bar\partial_{\dot\alpha}\) and \((\partial_\mu)^\dagger = - \partial_\mu\).

Starting from the differential operator realisation of \(P_\mu\), \(Q_\alpha\) and \(\bar Q_{\dot \alpha}\) \[P_\mu = -i\partial_\mu ~, \qquad Q_\alpha = - i \big(\partial_\alpha -i c (\sigma^\mu\bar\theta)_\alpha\partial_\mu\big) ~, \qquad \bar Q_{\dot\alpha} = i \big(\bar\partial_{\dot\alpha} - i \bar c (\theta \sigma^\mu)_{\dot\alpha}\partial_\mu\big) ~,\] where \(c\) and \(\bar c\) are free real constants, show that the anticommutation relations \[\{Q_\alpha,\bar Q_{\dot\alpha}\} = 2 (\sigma^{\mu})_{\alpha\dot\alpha} P_\mu ~, \qquad \{Q_\alpha, Q_{\beta}\} = \{\bar Q_{\dot\alpha}, \bar Q_{\dot\beta}\} = 0 ~,\] together with \((Q_\alpha)^\dagger = \bar Q_{\dot\alpha}\), imply that \(c=\bar c=1\). Show that the anticommutation relations are equivalent to \([\varepsilon_1 Q,\bar \varepsilon_2 \bar Q] = 2 \varepsilon_1\sigma^\mu\bar\varepsilon_2 P_\mu\) and \([\varepsilon_1 Q,\varepsilon_2 Q] = [\bar\varepsilon_1\bar Q,\bar\varepsilon_2\bar Q] = 0\).

Since \(Q_\alpha\) only depends on \(\partial_\alpha\) and \(\bar \theta^{\dot\alpha}\), while \(\bar Q_{\dot\alpha}\) only depends on \(\bar\partial_{\dot\alpha}\) and \(\theta^\alpha\), it immediately follows that \(\{Q_\alpha, Q_{\beta}\} = \{\bar Q_{\dot\alpha}, \bar Q_{\dot\beta}\} = 0\). For the final anticommutation relation we have \[\begin{split} \{Q_\alpha,\bar Q_{\dot\alpha}\} & = \{- i \big(\partial_\alpha -i c (\sigma^\mu\bar\theta)_\alpha\partial_\mu\big), i \big(\bar\partial_{\dot\alpha} - i \bar c (\theta \sigma^\mu)_{\dot\alpha}\partial_\mu\big)\} \\ & = \{\partial_\alpha,-i\bar c(\theta \sigma^\mu)_{\dot\alpha}\partial_\mu\} + \{-i c (\sigma^\mu\bar\theta)_\alpha\partial_\mu,\bar\partial_{\dot\alpha}\} \\ & = -i(\bar c+c) (\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu = (\bar c+c) (\sigma^\mu)_{\alpha\dot\alpha} P_\mu ~, \end{split}\] hence we require that \(\bar c+c = 2\). On the other hand \[\begin{split} (Q_\alpha)^\dagger & = i\big(\partial_\alpha -i c (\sigma^\mu\bar\theta)_\alpha\partial_\mu\big)^\dagger = i \big(\partial_{\dot\alpha} - i c (\bar\theta\sigma^\mu)_{\dot\alpha} \partial_\mu\big) ~, \end{split}\] where we have used that \(c\) is real. Therefore, \((Q_\alpha)^\dagger\) is equal to \(\bar Q_{\dot\alpha} = i \big(\bar\partial_{\dot\alpha} - i \bar c (\theta \sigma^\mu)_{\dot\alpha}\partial_\mu\big)\) if \(c = \bar c\). It immediately follows that \(c=\bar c= 1\).

Expanding out the commutator \([\varepsilon_1 Q,\bar \varepsilon_2 \bar Q]\) we find \[\begin{split} [\varepsilon_1 Q,\bar \varepsilon_2 \bar Q] & = - \varepsilon_1^\alpha Q_\alpha \varepsilon_2^{\dot \alpha} \bar Q_{\dot\alpha} + \varepsilon_2^{\dot\alpha} \bar Q_{\dot\alpha} \varepsilon_1^\alpha Q_\alpha = \varepsilon_1^\alpha\varepsilon_2^{\dot \alpha} Q_\alpha \bar Q_{\dot\alpha} + \varepsilon_1^\alpha\varepsilon_2^{\dot \alpha} \bar Q_{\dot\alpha} Q_\alpha \\ & = \varepsilon_1^\alpha\varepsilon_2^{\dot \alpha} \{Q_\alpha, \bar Q_{\dot\alpha} \} = 2 \varepsilon_1^\alpha\varepsilon_2^{\dot \alpha} (\sigma^\mu)_{\alpha\dot\alpha} P_\mu = 2 \varepsilon_1\sigma^\mu\bar\varepsilon_2 P_\mu ~. \end{split}\] Similarly, \[\begin{split} [\varepsilon_1 Q,\varepsilon_2 Q] & = - \varepsilon_1^\alpha Q_\alpha \varepsilon_2^{\beta} Q_{\beta} + \varepsilon_2^{\beta} Q_{\beta} \varepsilon_1^\alpha Q_\alpha = \varepsilon_1^\alpha\varepsilon_2^{\beta} Q_\alpha Q_{\beta} + \varepsilon_1^\alpha\varepsilon_2^{\beta} Q_{\beta} Q_\alpha = \varepsilon_1^\alpha\varepsilon_2^{\beta} \{Q_\alpha, Q_{\beta} \} = 0 ~, \\ [\bar \varepsilon_1 \bar Q,\bar \varepsilon_2 \bar Q] & = - \varepsilon_1^{\dot\alpha} \bar Q_{\dot\alpha} \varepsilon_2^{\dot \beta} \bar Q_{\dot\beta} + \varepsilon_2^{\dot\beta} \bar Q_{\dot\beta}\bar \varepsilon_1^{\dot\alpha} Q_{\dot\alpha} = \varepsilon_1^{\dot\alpha}\varepsilon_2^{\dot \beta} \bar Q_{\dot\alpha} \bar Q_{\dot\beta} + \varepsilon_1^{\dot\alpha}\varepsilon_2^{\dot \beta} \bar Q_{\dot\beta} \bar Q_{\dot\alpha} = \varepsilon_1^{\dot\alpha}\varepsilon_2^{\dot \beta} \{\bar Q_{\dot\alpha}, \bar Q_{\dot\beta} \} = 0 ~. \end{split}\]

Given that \(\{\partial_\alpha, \theta^\beta\} = \delta_\alpha^\beta\) and \(\{\bar\partial_{\dot\alpha},\bar\theta^{\dot\beta}\} = \delta_{\dot\alpha}^{\dot\beta}\), show that \(\{\partial^\alpha, \theta_\beta\} = -\delta^\alpha_\beta\) and \(\{\bar\partial^{\dot\alpha},\bar\theta_{\dot\beta}\} = -\delta^{\dot\alpha}_{\dot\beta}\), where we define \(\theta_\alpha = \epsilon_{\alpha\beta}\theta^\beta\), \(\partial^\alpha = \epsilon^{\alpha\beta}\partial_\beta\), \(\bar\theta_{\dot\alpha}=\epsilon_{\dot\alpha\dot\beta}\bar\theta^{\dot\beta}\) and \(\bar\partial^{\dot\alpha} = \epsilon^{\dot\alpha\dot\beta} \bar\partial_{\dot\beta}\), i.e. in the usual way.

We have \[\{\partial^\alpha, \theta_\beta\} = \epsilon^{\alpha\gamma}\epsilon_{\beta\delta}\{\partial_\gamma,\theta^\delta\} = \epsilon^{\alpha\gamma}\epsilon_{\beta\delta} \delta_\gamma^\delta = \epsilon^{\alpha\gamma}\epsilon_{\beta\gamma} = -\epsilon^{\alpha\gamma}\epsilon_{\gamma\beta} = -\delta^\alpha_\beta ~.\] Similarly \[\{\bar\partial^{\dot\alpha}, \bar\theta_{\dot\beta}\} = \epsilon^{\dot\alpha\dot\gamma}\epsilon_{\dot\beta\dot\delta}\{\bar\partial_{\dot\gamma},\bar\theta^{\dot\delta}\} = \epsilon^{\dot\alpha\dot\gamma}\epsilon_{\dot\beta\dot\delta} \delta_{\dot\gamma}^{\dot\delta} = \epsilon^{\dot\alpha\dot\gamma}\epsilon_{\dot\beta\dot\gamma} = -\epsilon^{\dot\alpha\dot\gamma}\epsilon_{\dot\gamma\dot\beta} = -\delta^{\dot\alpha}_{\dot\beta} ~.\]

**General superfield.**

A superfield is a field on superspace \(Y(x,\theta,\bar\theta)\). We can expand a general superfield as \[\begin{split} Y(x,\theta,\bar\theta) & = y(x) + \theta\psi(x) + \bar\theta \bar\chi(x) + (\theta\theta) m(x) + (\bar\theta\bar\theta)\bar n(x) \\ & \quad + (\theta\sigma^\mu\bar\theta) v_\mu(x) + (\theta\theta) (\bar\theta\bar\lambda(x)) + (\bar\theta\bar\theta) (\theta \rho(x)) + (\theta\theta)(\bar\theta\bar\theta) D(x) ~. \end{split}\] The Taylor expansion in \(\theta\) and \(\bar \theta\) terminates since \[\theta_\alpha\theta_\beta\theta_\gamma = \bar\theta_{\dot\alpha}\bar\theta_{\dot\beta}\bar\theta_{\dot\gamma} = 0 ~.\] This is a consequence of the fact that \(\theta_\alpha\) and \(\bar\theta_{\dot\alpha}\) are anticommuting coordinates, and that the indices \(\alpha = 1,2\) and \(\dot\alpha = 1,2\). The coefficients in the Taylor expansion are the component fields of the superfield from the component approach to off-shell supersymmetry discussed at the beginning of this section. Note that products and sums of superfields are also superfields.

Under infinitesimal translations and supersymmetry transformations a general superfield transforms as \[\begin{split} \delta_a Y(x,\theta,\bar\theta) = i [a^\mu P_\mu, Y(x,\theta,\bar\theta)] = a^\mu \partial_\mu Y(x,\theta,\bar\theta) ~, \qquad \delta_{\varepsilon,\bar\varepsilon} Y(x,\theta,\bar\theta) = i [\varepsilon Q + \bar\varepsilon \bar Q, Y(x,\theta,\bar\theta)] ~, \end{split}\] where \(Q_\alpha\) and \(\bar Q_{\dot\alpha}\) are realised in terms of differential operators \(\eqref{eq:qdiffop}\). Substituting in for \(Y(x,\theta,\bar\theta)\), \(Q_\alpha\) and \(\bar Q_{\dot\alpha}\) in \([\varepsilon Q + \bar\varepsilon \bar Q, Y(x,\theta,\bar\theta)]\), and equating with \[\delta_{\varepsilon,\bar\varepsilon} Y(x,\theta,\bar\theta) = \delta_{\varepsilon,\bar\varepsilon} y(x) + \theta \delta_{\varepsilon,\bar\varepsilon} \psi(x) + \bar\theta \delta_{\varepsilon,\bar\varepsilon} \bar\chi(x) + \dots + (\theta\theta)(\bar\theta\bar\theta) \delta_{\varepsilon,\bar\varepsilon} D(x) ~,\] we can extract the supersymmetry transformations of the component fields.

Count the number of bosonic and fermionic degrees of freedom of a general superfield \[\begin{split} Y(x,\theta,\bar\theta) & = y(x) + \theta\psi(x) + \bar\theta \bar\chi(x) + (\theta\theta) m(x) + (\bar\theta\bar\theta)\bar n(x) \\ & \quad + (\theta\sigma^\mu\bar\theta) v_\mu(x) + (\theta\theta) (\bar\theta\bar\lambda(x)) + (\bar\theta\bar\theta) (\theta \rho(x)) + (\theta\theta)(\bar\theta\bar\theta) D(x) ~. \end{split}\] Derive the supersymmetry transformations of the component fields by equating \[\delta_{\varepsilon,\bar\varepsilon} Y(x,\theta,\bar\theta) = i [\varepsilon Q + \bar\varepsilon \bar Q,Y(x,\theta,\bar\theta)] ~,\] and \[\delta_{\varepsilon,\bar\varepsilon} Y(x,\theta,\bar\theta) = \delta_{\varepsilon,\bar\varepsilon} y(x) + \theta \delta_{\varepsilon,\bar\varepsilon} \psi(x) + \bar\theta \delta_{\varepsilon,\bar\varepsilon} \bar\chi(x) + \dots + (\theta\theta)(\bar\theta\bar\theta) \delta_{\varepsilon,\bar\varepsilon} D(x) ~,\] where \[Q_\alpha = - i \big(\partial_\alpha -i(\sigma^\mu\bar\theta)_\alpha\partial_\mu\big) ~, \qquad \bar Q_{\dot\alpha} = i \big(\bar\partial_{\dot\alpha} - i (\theta \sigma^\mu)_{\dot\alpha}\partial_\mu\big) ~.\] What is special about the variation of \(D\)?

The bosonic component fields are the complex scalars \(y\), \(m\), \(\bar n\) and \(D\) each with 2 real degrees of freedom, and the complex vector \(v_\mu\) with \(2\times 4 = 8\) real degrees of freedom. Therefore, the total number of real bosonic degrees of freedom is 16. The fermionic component fields are the Weyl fermions \(\psi\), \(\bar\chi\), \(\bar\lambda\) and \(\rho\) each with 4 real degrees of freedom. Therefore, the total number of real fermionic degrees of freedom is also 16.

To derive the supersymmetry transformations of the component fields, we start by noting that \[i (\varepsilon Q + \bar\varepsilon \bar Q) = \varepsilon^{\alpha}\partial_\alpha - i(\varepsilon\sigma^\mu\bar\theta)\partial_\mu + \bar\varepsilon^{\dot\alpha} \bar\partial_{\dot\alpha} + i (\theta \sigma^\mu\bar\varepsilon)\partial_\mu ~,\] and observe that \(\varepsilon^{\alpha}\partial_\alpha\) lowers the power of \(\theta\) by one, while \(i (\theta \sigma^\mu\bar\varepsilon)\partial_\mu\) raises it by one. Similarly, \(\bar\varepsilon^{\dot\alpha} \bar\partial_{\dot\alpha}\) lowers the power of \(\bar \theta\) by one, while \(- i(\varepsilon\sigma^\mu\bar\theta)\partial_\mu\) raises it by one.

To derive the supersymmetry transformation of \(y\), i.e. \(\delta_{\varepsilon,\bar\varepsilon} y\), we need to extract those terms in \(i [\varepsilon Q + \bar\varepsilon \bar Q, Y(x,\theta,\bar\theta)]\) that are independent of \(\theta\) and \(\bar\theta\). These terms can only come from lowering the power of \(\theta\) in \(\theta \psi\) by one, i.e. applying \(\varepsilon^{\alpha}\partial_\alpha\), or lowering the power of \(\bar\theta\) in \(\bar\theta \bar\chi\) by one, i.e. applying \(\bar\varepsilon^{\dot\alpha} \bar\partial_{\dot\alpha}\). Therefore, it follows that \[\delta_{\varepsilon,\bar\varepsilon} y = \varepsilon \psi + \bar\varepsilon \bar\chi ~.\]

To derive the supersymmetry transformation of \(\psi\), i.e. \(\delta_{\varepsilon,\bar\varepsilon} \psi\), we need to extract those terms in \(i [\varepsilon Q + \bar\varepsilon \bar Q, Y(x,\theta,\bar\theta)]\) that are linear in \(\theta\) and independent of \(\bar\theta\). There are three such terms, which come from raising the power of \(\theta\) in \(y\), lowering the power of \(\theta\) in \((\theta\theta)m\) by one, and lowering the power of \(\bar\theta\) in \((\theta\sigma^\mu\bar\theta)v_\mu\) by one. Therefore, we have \[\begin{split} \theta \delta_{\varepsilon,\bar\varepsilon} \psi & = i (\theta \sigma^\mu\bar\varepsilon)\partial_\mu y + 2 (\varepsilon\theta) m + (\theta\sigma^\mu\bar\varepsilon) v_\mu = i (\theta \sigma^\mu\bar\varepsilon)\partial_\mu y + 2 (\theta\varepsilon) m + (\theta\sigma^\mu\bar\varepsilon) v_\mu ~, \end{split}\] which implies \[\delta_{\varepsilon,\bar\varepsilon} \psi = i (\sigma^\mu\bar\varepsilon)\partial_\mu y + 2 \varepsilon m + (\sigma^\mu\bar\varepsilon) v_\mu ~.\]

For the supersymmetry transformation of \(\bar\chi\), i.e. \(\delta_{\varepsilon,\bar\varepsilon} \bar\chi\), we have \[\begin{split} \bar \theta \delta_{\varepsilon,\bar\varepsilon} \bar\chi & = - i(\varepsilon\sigma^\mu\bar\theta)\partial_\mu y + 2(\bar\varepsilon\bar\theta)\bar n + (\varepsilon\sigma^\mu\bar\theta)v_\mu = i (\bar\theta\bar\sigma^\mu\varepsilon)\partial_\mu y + 2(\bar\theta\bar\varepsilon)\bar n -(\bar\theta\bar\sigma^\mu\varepsilon)v_\mu ~, \end{split}\] which implies \[\delta_{\varepsilon,\bar\varepsilon} \bar\chi = i (\bar\sigma^\mu\varepsilon)\partial_\mu y + 2\bar\varepsilon\bar n -(\bar\sigma^\mu\varepsilon)v_\mu ~.\]

For the supersymmetry transformation of \(m\), i.e. \(\delta_{\varepsilon,\bar\varepsilon}m\), we have \[\begin{split} (\theta\theta) \delta_{\varepsilon,\bar\varepsilon}m & = i (\theta \sigma^\mu\bar\varepsilon)(\theta\partial_\mu\psi) + (\theta\theta)(\bar\varepsilon\bar\lambda) = - \frac{i}{2}(\theta\theta)(\partial_\mu\psi\sigma^\mu\bar\varepsilon) + (\theta\theta)(\bar\varepsilon\bar\lambda) \\ & = \frac{i}{2}(\theta\theta)(\bar\varepsilon\bar\sigma^\mu\partial_\mu\psi) + (\theta\theta)(\bar\varepsilon\bar\lambda) ~, \end{split}\] which implies \[\delta_{\varepsilon,\bar\varepsilon}m = \frac{i}{2}\bar\varepsilon\bar\sigma^\mu\partial_\mu\psi + \bar\varepsilon\bar\lambda ~.\]

For the supersymmetry transformation of \(\bar n\), i.e. \(\delta_{\varepsilon,\bar\varepsilon}\bar n\), we have \[\begin{split} (\bar\theta\bar\theta) \delta_{\varepsilon,\bar\varepsilon}\bar n & = - i(\varepsilon\sigma^\mu\bar\theta) (\bar\theta\partial_\mu\bar\chi) + (\bar\theta\bar\theta)(\varepsilon\rho) = i(\bar\theta\bar\sigma^\mu\varepsilon) (\bar\theta\partial_\mu\bar\chi) + (\bar\theta\bar\theta)(\varepsilon\rho) \\ & = -\frac{i}{2}(\bar\theta\bar\theta)(\partial_\mu\bar\chi\bar\sigma^\mu\varepsilon) + (\bar\theta\bar\theta)(\varepsilon\rho) = \frac{i}{2}(\bar\theta\bar\theta)(\varepsilon\sigma^\mu\partial_\mu\bar\chi) + (\bar\theta\bar\theta)(\varepsilon\rho) ~, \end{split}\] which implies \[\delta_{\varepsilon,\bar\varepsilon}\bar n = \frac{i}{2}\varepsilon\sigma^\mu\partial_\mu\bar\chi + \varepsilon\rho ~.\]

For the supersymmetry transformation of \(v_\mu\), i.e. \(\delta_{\varepsilon,\bar\varepsilon}v_\mu\), we have \[\begin{split} (\theta\sigma^\mu\bar\theta)\delta_{\varepsilon,\bar\varepsilon}v_\mu & = - i(\varepsilon\sigma^\nu\bar\theta)(\theta\partial_\nu\psi) + i (\theta \sigma^\nu\bar\varepsilon)(\bar\theta\partial_\nu\bar\chi) +2 (\varepsilon\theta)(\bar\theta\bar\lambda) +2 (\bar\varepsilon\bar\theta)(\theta\rho) \\ & = i(\bar\theta\bar\sigma^\nu\varepsilon)(\theta\partial_\nu\psi) + i (\theta \sigma^\nu\bar\varepsilon)(\bar\theta\partial_\nu\bar\chi) +2 (\theta\varepsilon)(\bar\theta\bar\lambda) +2 (\bar\theta\bar\varepsilon)(\theta\rho) \\ & = \frac{i}{2}(\theta\sigma^\mu\bar\theta)(\partial_\nu\psi\sigma_\mu\bar\sigma^\nu\varepsilon) - \frac{i}{2}(\theta\sigma^\mu\bar\theta)(\partial_\nu\bar\chi\bar\sigma_\mu\sigma^\nu\bar\varepsilon) +(\theta\sigma^\mu\bar\theta) (\varepsilon\sigma_\mu \bar\lambda) +(\theta\sigma^\mu\bar\theta)(\rho\sigma_\mu\bar\varepsilon) \\ & = \frac{i}{2}(\theta\sigma^\mu\bar\theta)(\varepsilon\sigma^\nu\bar\sigma_\mu\partial_\nu\psi) - \frac{i}{2}(\theta\sigma^\mu\bar\theta)(\bar\varepsilon\bar\sigma^\nu\sigma_\mu\partial_\nu\bar\chi) +(\theta\sigma^\mu\bar\theta) (\varepsilon\sigma_\mu \bar\lambda) -(\theta\sigma^\mu\bar\theta)(\bar\varepsilon\bar\sigma_\mu\rho) ~, \end{split}\] which implies \[\delta_{\varepsilon,\bar\varepsilon}v_\mu = \frac{i}{2}\varepsilon\sigma^\nu\bar\sigma_\mu\partial_\nu\psi - \frac{i}{2}\bar\varepsilon\bar\sigma^\nu\sigma_\mu\partial_\nu\bar\chi +\varepsilon\sigma_\mu \bar\lambda -\bar\varepsilon\bar\sigma_\mu\rho ~.\]

For the supersymmetry transformation of \(\bar\lambda\), i.e. \(\delta_{\varepsilon,\bar\varepsilon}\bar\lambda\), we have \[\begin{split} (\theta\theta)(\bar\theta\delta_{\varepsilon,\bar\varepsilon}\bar\lambda) & = -i(\varepsilon\sigma^\mu\bar\theta)(\theta\theta)\partial_\mu m +i (\theta \sigma^\mu\bar\varepsilon)(\theta\sigma^\nu\bar\theta)\partial_\mu v_\nu +2(\theta\theta)(\bar\varepsilon\bar\theta)D \\ & = i(\theta\theta)(\bar\theta\bar\sigma^\mu\varepsilon)\partial_\mu m +\frac{i}{2}(\theta\theta)(\bar\theta\bar\sigma^\nu\sigma^\mu\bar\varepsilon)\partial_\mu v_\nu +2(\theta\theta)(\bar\theta\bar\varepsilon)D ~, \end{split}\] which implies \[\delta_{\varepsilon,\bar\varepsilon}\bar\lambda = i(\bar\sigma^\mu\varepsilon)\partial_\mu m +\frac{i}{2}(\bar\sigma^\nu\sigma^\mu\bar\varepsilon)\partial_\mu v_\nu +2\bar\varepsilon D ~.\]

For the supersymmetry transformation of \(\rho\), i.e. \(\delta_{\varepsilon,\bar\varepsilon}\rho\), we have \[\begin{split} (\bar\theta\bar\theta)(\theta\delta_{\varepsilon,\bar\varepsilon}\rho) & = i (\theta \sigma^\mu\bar\varepsilon)(\bar\theta\bar\theta)\partial_\mu \bar n - i(\varepsilon\sigma^\mu\bar\theta)(\theta\sigma^\nu\bar\theta)\partial_\mu v_\nu +2(\varepsilon\theta)(\bar\theta\bar\theta)D \\ & = i (\bar\theta\bar\theta)(\theta \sigma^\mu\bar\varepsilon)\partial_\mu \bar n -i(\bar\theta\bar\sigma^\mu\varepsilon)(\bar\theta\bar\sigma^\nu\theta)\partial_\mu v_\nu +2(\bar\theta\bar\theta)(\theta\varepsilon)D \\ & = i (\bar\theta\bar\theta)(\theta \sigma^\mu\bar\varepsilon)\partial_\mu \bar n -\frac{i}{2}(\bar\theta\bar\theta)(\theta\sigma^\nu\bar\sigma^\mu\varepsilon)\partial_\mu v_\nu +2(\bar\theta\bar\theta)(\theta\varepsilon)D ~, \end{split}\] which implies \[\delta_{\varepsilon,\bar\varepsilon}\rho = i (\sigma^\mu\bar\varepsilon)\partial_\mu \bar n -\frac{i}{2}(\sigma^\nu\bar\sigma^\mu\varepsilon)\partial_\mu v_\nu +2\varepsilon D ~,\]

For the supersymmetry transformation of \(D\), i.e. \(\delta_{\varepsilon,\bar\varepsilon} D\), we have \[\begin{split} (\theta\theta)(\bar\theta\bar\theta)\delta_{\varepsilon,\bar\varepsilon} D & = - i(\varepsilon\sigma^\mu\bar\theta)(\theta\theta)(\bar\theta\partial_\mu\bar\lambda) + i (\theta \sigma^\mu\bar\varepsilon)(\bar\theta\bar\theta)(\theta\partial_\mu\rho) \\ & = i(\theta\theta) (\bar\theta\bar\sigma^\mu\varepsilon)(\bar\theta\partial_\mu\bar\lambda) + i (\bar\theta\bar\theta)(\theta \sigma^\mu\bar\varepsilon)(\theta\partial_\mu\rho) \\ & = -\frac{i}{2}(\theta\theta)(\bar\theta\bar\theta)(\partial_\mu\bar\lambda\bar\sigma^\mu\varepsilon) - \frac{i}{2}(\theta\theta)(\bar\theta\bar\theta)(\partial_\mu\rho\sigma^\mu\bar\varepsilon) \\ & = \frac{i}{2}(\theta\theta)(\bar\theta\bar\theta)(\varepsilon\sigma^\mu\partial_\mu\bar\lambda) +\frac{i}{2}(\theta\theta)(\bar\theta\bar\theta)(\bar\varepsilon\bar\sigma^\mu\partial_\mu\rho) ~, \end{split}\] which implies \[\delta_{\varepsilon,\bar\varepsilon} D = \frac{i}{2}\varepsilon\sigma^\mu\partial_\mu\bar\lambda +\frac{i}{2}\bar\varepsilon\bar\sigma^\mu\partial_\mu\rho ~,\] which is a total derivative.

Note that we have made repeated use of the spinor identities derived in exercise 2.4.

A general superfield \(Y(x,\theta,\bar\theta)\) contains too many component fields to be an irreducible representation of the supersymmetry algebra. Indeed, we can reduce the number of components by imposing a constraint that is covariant under supersymmetry, i.e. a supersymmetry transformation does not reintroduce the components that the constraint eliminated. ^{17} One way to introduce a covariant constraint, which we will return to in section 6, is to impose a reality condition \(Y = Y^\dagger\). Another way, which we will discuss now, is to impose a differential constraint on superspace. The resulting superfields will be the subject of section 5.

**Chiral superfields.**

In order to impose a differential constraint that is covariant under supersymmetry, we define supercovariant derivatives \(D_\alpha\) and \(\bar D_{\dot\alpha}\) that anticommute with the supercharges (realised as differential operators) \[D_\alpha = \partial_\alpha + i (\sigma^\mu\bar\theta)_\alpha \partial_\mu ~, \qquad \bar D_{\dot\alpha} = \bar\partial_{\dot\alpha} + i (\theta\sigma^\mu)_{\dot\alpha} \partial_\mu ~.\] These differential operators satisfy the following relations \[\begin{gathered} \{D_\alpha,Q_\beta\} = \{D_\alpha,\bar Q_{\dot\alpha}\} = \{\bar D_{\dot\alpha},Q_\alpha\} = \{\bar D_{\dot\alpha},\bar Q_{\dot\beta} \} = 0 ~, \\ \{D_\alpha,\bar D_{\dot\alpha}\} = 2 i (\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu ~, \qquad \{D_\alpha,D_\beta\} = \{\bar D_{\dot\alpha},\bar D_{\dot\beta} \} = 0 ~. \end{gathered}\] We also have the conjugation relations \((D_\alpha)^\dagger = \bar D_{\dot\alpha}\).

Starting from the differential operators \[\begin{aligned} Q_\alpha & = -i\big( \partial_\alpha - i (\sigma^\mu\bar\theta)_\alpha \partial_\mu \big)~, \qquad & \bar Q_{\dot\alpha} & = i\big(\bar\partial_{\dot\alpha} - i (\theta\sigma^\mu)_{\dot\alpha} \partial_\mu \big) ~, \\ D_\alpha & = \partial_\alpha + i (\sigma^\mu\bar\theta)_\alpha \partial_\mu ~, \qquad & \bar D_{\dot\alpha} & = \bar\partial_{\dot\alpha} + i (\theta\sigma^\mu)_{\dot\alpha} \partial_\mu ~, \end{aligned}\] show that \[\begin{gathered} \{D_\alpha,Q_\beta\} = \{D_\alpha,\bar Q_{\dot\alpha}\} = \{\bar D_{\dot\alpha},Q_\alpha\} = \{\bar D_{\dot\alpha},\bar Q_{\dot\beta} \} = 0 ~, \\ \{D_\alpha,\bar D_{\dot\alpha}\} = 2 i (\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu ~, \qquad \{D_\alpha,D_\beta\} = \{\bar D_{\dot\alpha},\bar D_{\dot\beta} \} = 0 ~. \end{gathered}\]

Since \(Q_\alpha\) and \(D_\alpha\) only depend on \(\partial_\alpha\) and \(\bar \theta^{\dot\alpha}\), while \(\bar Q_{\dot\alpha}\) and \(\bar D_{\dot\alpha}\) only depend on \(\bar\partial_{\dot\alpha}\) and \(\theta^\alpha\), it immediately follows that \(\{D_\alpha, Q_{\beta}\} = \{\bar D_{\dot\alpha}, \bar Q_{\dot\beta}\} = \{D_\alpha, D_{\beta}\} = \{\bar D_{\dot\alpha}, \bar D_{\dot\beta}\} = 0\). For the remaining three anticommutation relations we have \[\begin{split} \{D_\alpha,\bar Q_{\dot\alpha}\} & = \{\partial_\alpha,(\theta\sigma^\mu)_{\dot\alpha}\partial_\mu\} + \{i(\sigma^\mu\bar\theta)_\alpha\partial_\mu,i\bar\partial_{\dot\alpha}\} =(\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu - (\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu = 0 ~, \\ \{\bar D_{\dot\alpha},Q_{\alpha}\} & = \{\bar\partial_{\dot\alpha},-(\sigma^\mu\bar\theta)_{\alpha}\partial_\mu\} + \{i(\theta\sigma^\mu)_{\dot\alpha}\partial_\mu,-i\partial_{\alpha}\} =-(\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu + (\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu = 0 ~, \\ \{D_\alpha,\bar D_{\dot\alpha}\} & = \{\partial_\alpha,i(\theta\sigma^\mu)_{\dot\alpha}\partial_\mu\} + \{i(\sigma^\mu\bar\theta)_\alpha\partial_\mu,\bar\partial_{\dot\alpha}\} =2i(\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu ~. \end{split}\]

Since the supercovariant derivatives anticommute with the supercharges, they can be used to impose differential constraints that are covariant under supersymmetry. As an explicit example consider a superfield \(\Phi\) constrained such that \(\bar D_{\dot\alpha} \Phi = 0\). Under supersymmetry transformations \[\Phi \to \tilde \Phi = e^{i (\varepsilon Q + \bar \varepsilon \bar Q)} \Phi e^{-i(\varepsilon Q + \bar \varepsilon \bar Q)} ~.\] Since \(\bar D_{\dot\alpha}\) anticommutes with both \(Q_\alpha\) and \(\bar Q_{\dot\alpha}\) the transformed superfield \(\tilde \Phi\) also satisfies \(\bar D_{\dot\alpha} \tilde \Phi = 0\) and no new degrees of freedom have been introduced. We say that the differential constraint \(\bar D_{\dot\alpha} \Phi = 0\) is covariant under supersymmetry.

A superfield \(\Phi\) constrained such that \(\bar D_{\dot\alpha} \Phi = 0\) is known as a chiral superfield. A superfield \(\bar\Phi\) constrained such that \(D_\alpha \bar \Phi = 0\) is known as an antichiral superfield. The conjugate of a chiral superfield is an antichiral superfield. Therefore, we write \(\bar \Phi = (\Phi)^\dagger\). If \(\Phi\) is both chiral and antichiral then it must be constant since \(0 = \{D_\alpha,\bar D_{\dot\alpha}\} \Phi = 2 i (\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu \Phi\) implies \(\partial_\mu \Phi = 0\). To solve the chiral and antichiral constraints we introduce the chiral and antichiral superspace coordinates \[y^\mu = x^\mu + i \theta \sigma^\mu\bar \theta ~, \qquad \bar y^\mu = x^\mu - i \theta \sigma^\mu \bar \theta ~,\] where \(\bar y^\mu = (y^\mu)^\dagger\). These are defined such that \[\begin{aligned} \{D_\alpha,\theta^\beta\} & = \delta_{\alpha}^{\beta} ~, & \qquad \{\bar D_{\dot\alpha},\theta^\alpha\} & = \bar [D_{\dot\alpha}, y^\mu] = 0 ~, \\ \{\bar D_{\dot\alpha},\bar\theta^{\dot\beta}\} & = \delta_{\dot\alpha}^{\dot\beta} ~, & \qquad \{D_{\alpha},\bar\theta^{\dot\alpha}\} & = [D_{\alpha}, \bar y^\mu] = 0 ~. \end{aligned}\]

Show that \[= [D_{\alpha}, \bar y^\mu] = 0 ~,\] where \[\begin{aligned} y^\mu & = x^\mu + i \theta \sigma^\mu\bar \theta ~, & \qquad D_\alpha & = \partial_\alpha + i (\sigma^\mu\bar\theta)_\alpha \partial_\mu ~, \\ \bar y^\mu & = x^\mu - i \theta \sigma^\mu \bar \theta ~, & \qquad \bar D_{\dot\alpha} & = \bar\partial_{\dot\alpha} + i (\theta\sigma^\mu)_{\dot\alpha} \partial_\mu ~. \end{aligned}\]

We have \[= [\bar\partial_{\dot\alpha} + i (\theta\sigma^\nu)_{\dot\alpha} \partial_\nu,x^\mu + i \theta \sigma^\mu \bar \theta] = - i (\theta\sigma^\mu)_{\dot\alpha} + i(\theta\sigma^\mu)_{\dot\alpha} = 0 ~,\] and \[= [\partial_{\alpha} + i (\sigma^\nu\bar\theta)_{\alpha} \partial_\nu,x^\mu - i \theta \sigma^\mu \bar \theta] = - i (\sigma^\mu\bar\theta)_{\alpha} + i(\sigma^\mu\bar\theta)_{\alpha} = 0 ~.\]

Let us now solve the chiral constraint and derive the supersymmetry transformations of the component fields of a chiral superfield. An analogous derivation holds for antichiral superfields. The chiral constraint \(\bar D_{\dot\alpha}\Phi=0\) is solved by \[\begin{equation} \label{eq:chiral} \Phi(y,\theta) = \phi(y) + \sqrt{2} \theta \psi(y) - (\theta\theta) F(y) ~. \end{equation}\] Written in this form, it is clear that products and sums of chiral superfields are also chiral superfields. The former also follows from the Leibniz rule. Taylor expanding \(\eqref{eq:chiral}\) around \(y^\mu = x^\mu\), a chiral superfield expressed in terms of the original coordinates \((x,\theta,\bar\theta)\) is \[\begin{split} \Phi(x,\theta,\bar\theta) & = \phi(x) + i (\theta\sigma^\mu\bar\theta)\partial_\mu\phi(x) - \frac{1}{4}(\theta\theta)(\bar\theta\bar\theta) \partial_\mu\partial^\mu \phi(x) \\ & \quad + \sqrt{2}\theta\psi(x) -\frac{i}{\sqrt{2}}(\theta\theta)(\partial_\mu\psi(x) \sigma^\mu\bar\theta) -(\theta\theta) F(x) ~, \end{split}\] where we have used the spinor identities \[\begin{equation} \label{eq:spinident} (\theta\phi) (\theta\psi) = -\frac{1}{2} (\theta\theta)(\phi\psi) ~, \qquad (\theta \sigma^\mu\bar\theta)(\theta\sigma^\nu\bar\theta) = \frac{1}{2}(\theta\theta)(\bar\theta\bar\theta) \eta^{\mu\nu} ~, \end{equation}\] which were proven in section 2.

To find the supersymmetry transformation of a chiral superfield, it is useful to change coordinates from \((x,\theta,\bar\theta)\) to \((y,\theta,\bar\theta)\). Denoting the partial derivatives with respect to \((y,\theta,\bar\theta)\) by \((\hat \partial_\mu,\hat\partial_\alpha,\hat{\bar\partial}_{\dot\alpha})\) we find \[Q_\alpha = -i \hat\partial_\alpha ~, \qquad \bar Q_{\dot\alpha} = i\big(\hat{\bar\partial}_{\dot\alpha} - 2 i (\theta\sigma^\mu)_{\dot\alpha} \hat\partial_{\mu}\big) ~.\] Therefore, \[\begin{split} \delta_{\varepsilon,\bar\varepsilon} \Phi(y,\theta) & = i[\varepsilon Q+\bar \varepsilon \bar Q,\Phi(y,\theta)] = \big(\varepsilon^\alpha\hat\partial_\alpha +2i(\theta \sigma^\mu\bar\varepsilon)\hat\partial_\mu \big)\Phi(y,\theta) \\ & = \big(\varepsilon^\alpha\hat\partial_\alpha +2i(\theta \sigma^\mu\bar\varepsilon)\hat\partial_\mu \big) \big(\phi(y) + \sqrt{2} \theta \psi(y) - (\theta\theta) F(y)\big) \\ & = \sqrt{2}\varepsilon\psi(y) + 2i(\theta \sigma^\mu\bar\varepsilon)\hat\partial_\mu \phi(y) - 2 (\varepsilon\theta) F(y) + 2\sqrt{2}i(\theta \sigma^\mu\bar\varepsilon)(\theta\hat\partial_\mu \psi(y)) \\ & = \sqrt{2}\varepsilon\psi(y) + 2i(\theta \sigma^\mu\bar\varepsilon)\hat\partial_\mu \phi(y) - 2 (\theta\varepsilon) F(y) - \sqrt{2}i(\theta\theta)(\hat\partial_\mu \psi(y) \sigma^\mu\bar\varepsilon)~, \end{split}\] where we have used the first of the spinor identities \(\eqref{eq:spinident}\). Equating this expression with \[\delta_{\varepsilon,\bar\varepsilon} \Phi(y,\theta) =\delta_{\varepsilon,\bar\varepsilon} \phi(y) + \sqrt{2} \theta \delta_{\varepsilon,\bar\varepsilon} \psi(y) - (\theta\theta)\delta_{\varepsilon,\bar\varepsilon} F(y) ~,\] we can read off the supersymmetry transformations of the component fields of a chiral superfield \[\begin{split} \delta_{\varepsilon,\bar\varepsilon} \phi & = \sqrt{2}\varepsilon\psi ~, \\ \delta_{\varepsilon,\bar\varepsilon} \psi & = \sqrt{2}i(\sigma^\mu\bar\varepsilon)\partial_\mu \phi - \sqrt{2}\varepsilon F ~, \\ \delta_{\varepsilon,\bar\varepsilon} F & = \sqrt{2}i\partial_\mu \psi\sigma^\mu\bar\varepsilon ~. \end{split}\]

Express the supercovariant derivatives \(D_\alpha\) and \(\bar D_{\dot\alpha}\) and the supercharges \(Q_\alpha\) and \(\bar Q_{\dot\alpha}\), realised as differential operators, \[\begin{aligned} D_\alpha & = \partial_\alpha + i (\sigma^\mu\bar\theta)_\alpha \partial_\mu ~, \qquad & \bar D_{\dot\alpha} & = \bar\partial_{\dot\alpha} + i (\theta\sigma^\mu)_{\dot\alpha} \partial_\mu ~, \\ Q_\alpha & = -i\big( \partial_\alpha - i (\sigma^\mu\bar\theta)_\alpha \partial_\mu \big)~, \qquad & \bar Q_{\dot\alpha} & = i\big(\bar\partial_{\dot\alpha} - i (\theta\sigma^\mu)_{\dot\alpha} \partial_\mu \big) ~, \end{aligned}\] in terms of derivatives with respect to \((y,\theta,\bar\theta)\) and \((\bar y,\theta,\bar\theta)\) where \[y^\mu = x^\mu + i \theta \sigma^\mu\bar \theta ~, \qquad \bar y^\mu = x^\mu - i \theta \sigma^\mu \bar \theta ~.\]

We introduce the notation \((\hat \partial_\mu,\hat\partial_\alpha,\hat{\bar\partial}_{\dot\alpha})\) to denote partial derivatives with respect to \((y,\theta,\bar\theta)\) and continue to use \((\partial_\mu,\partial_\alpha,\bar\partial_{\dot\alpha})\) to denote partial derivatives with respect to \((x,\theta,\bar\theta)\). We then have \[\begin{split} \partial_\mu & = \frac{\partial y^\nu}{\partial x^\mu} \hat\partial_{\nu} + \frac{\partial \theta^\alpha}{\partial x^\mu} \hat\partial_\alpha + \frac{\partial \bar\theta^{\dot\alpha}}{\partial x^\mu} \hat{\bar\partial}_{\dot\alpha} = \delta_\mu^\nu \hat\partial_{\nu} = \hat\partial_{\mu} ~, \\ \partial_\alpha & = \frac{\partial y^\mu}{\partial \theta^\alpha} \hat\partial_{\mu} + \frac{\partial \theta^\beta}{\partial \theta^\alpha} \hat\partial_\beta + \frac{\partial \bar\theta^{\dot\alpha}}{\partial \theta^\alpha} \hat{\bar\partial}_{\dot\alpha} = i (\sigma^\mu\bar\theta)_{\alpha} \hat\partial_{\mu} + \delta_{\alpha}^\beta \hat\partial_\beta = \hat\partial_\alpha + i (\sigma^\mu\bar\theta)_{\alpha} \hat\partial_{\mu} ~, \\ \bar\partial_{\dot\alpha} & = \frac{\partial y^\mu}{\partial \bar\theta^{\dot\alpha}} \hat\partial_{\mu} + \frac{\partial \theta^\alpha}{\partial \bar\theta^{\dot\alpha}} \hat\partial_\alpha + \frac{\partial \bar\theta^{\dot\beta}}{\partial \bar \theta^{\dot\alpha}} \hat{\bar\partial}_{\dot\beta} = - i (\theta\sigma^\mu)_{\dot\alpha} \hat\partial_{\mu} + \delta_{\dot\alpha}^{\dot\beta} \hat{\bar\partial}_{\dot\beta} = \hat{\bar\partial}_{\dot\alpha} - i (\theta\sigma^\mu)_{\dot\alpha} \hat\partial_{\mu} ~. \end{split}\] Substituting into the expressions for the supercovariant derivatives and supercharges realised as differential operators we find \[\begin{aligned} D_\alpha & = \hat\partial_\alpha + 2i (\sigma^\mu\bar\theta)_{\alpha} \hat\partial_{\mu} ~, \qquad & \bar D_{\dot\alpha} & = \hat{\bar\partial}_{\dot\alpha} ~, \\ Q_\alpha & = -i \hat\partial_\alpha ~, & \qquad \bar Q_{\dot\alpha} & = i\big(\hat{\bar\partial}_{\dot\alpha} - 2 i (\theta\sigma^\mu)_{\dot\alpha} \hat\partial_{\mu}\big) ~. \end{aligned}\]

A similar derivation shows that \[\begin{aligned} D_\alpha & = \tilde\partial_\alpha ~, \qquad & \bar D_{\dot\alpha} & = \tilde{\bar\partial}_{\dot\alpha} + 2 i (\theta\sigma^\mu)_{\dot\alpha} \tilde\partial_{\mu} ~, \\ Q_\alpha & = -i \big(\tilde\partial_\alpha - 2i (\sigma^\mu\bar\theta)_{\alpha} \tilde\partial_{\mu}\big) ~, & \qquad \bar Q_{\dot\alpha} & = i\tilde{\bar\partial}_{\dot\alpha} ~, \end{aligned}\] where \((\tilde \partial_\mu,\tilde\partial_\alpha,\tilde{\bar\partial}_{\dot\alpha})\) denote partial derivatives with respect to \((\bar y,\theta,\bar\theta)\).

Starting from a chiral superfield expressed in terms of components as \[\Phi(y,\theta) = \phi(y) + \sqrt{2} \theta \psi(y) - (\theta\theta) F(y) ~,\] derive the component field expansion in terms of the coordinates \((x,\theta,\bar\theta)\) \[\begin{split} \Phi(x,\theta,\bar\theta) & = \phi(x) + i (\theta\sigma^\mu\bar\theta)\partial_\mu\phi(x) - \frac{1}{4}(\theta\theta)(\bar\theta\bar\theta) \partial_\mu\partial^\mu \phi(x) \\ & \quad + \sqrt{2}\theta\psi(x) -\frac{i}{\sqrt{2}}(\theta\theta)(\partial_\mu\psi(x) \sigma^\mu\bar\theta) -(\theta\theta) F(x) ~, \end{split}\] where \(y^\mu = x^\mu + i \theta \sigma^\mu\bar \theta\).

We start by Taylor expanding \(\Phi(y,\theta)\) around \(y^\mu = x^\mu\) to give \[\Phi(x,\theta,\bar\theta) = \Phi(x,\theta) + i (\theta \sigma^\mu\bar \theta) \partial_\mu \Phi(x,\theta) - \frac{1}{2} (\theta \sigma^\mu\bar \theta)(\theta \sigma^\nu\bar \theta) \partial_\mu\partial_\nu \Phi(x,\theta) ~.\] The expansion terminates at quadratic order in either \(\theta_{\alpha}\) or \(\bar\theta_{\dot\alpha}\) since they are anticommuting coordinates and the indices \(\alpha=1,2\) and \(\dot\alpha=1,2\). Now substituting in for \(\Phi(x,\theta)\) we find \[\begin{split} \Phi(x,\theta,\bar\theta) & = \phi(x) + \sqrt{2} \theta \psi(x) - (\theta\theta) F(x) \\ & \quad + i (\theta \sigma^\mu\bar \theta) \partial_\mu \phi(x) + \sqrt{2} i (\theta \sigma^\mu\bar \theta) (\theta \partial_\mu\psi(x)) - \frac{1}{2} (\theta \sigma^\mu\bar \theta)(\theta \sigma^\nu\bar \theta) \partial_\mu\partial_\nu \phi(x) ~. \end{split}\] Finally, using the following spinor identities \[(\theta\phi) (\theta\psi) = -\frac{1}{2} (\theta\theta)(\phi\psi) ~, \qquad (\theta \sigma^\mu\bar\theta)(\theta\sigma^\nu\bar\theta) = \frac{1}{2}(\theta\theta)(\bar\theta\bar\theta) \eta^{\mu\nu} ~,\] which were proven in exercise 2.4, we can rearrange the final two terms to give \[\begin{split} \Phi(x,\theta,\bar\theta) & = \phi(x) + i (\theta \sigma^\mu\bar \theta) \partial_\mu \phi(x) - \frac{1}{4}(\theta\theta)(\bar\theta\bar\theta) \partial_\mu\partial^\mu \phi(x) \\ & \quad + \sqrt{2} \theta \psi(x) -\frac{i}{\sqrt{2}} (\theta \theta)( \partial_\mu\psi(x) \sigma^\mu\bar \theta) - (\theta\theta) F(x) ~, \end{split}\] as required.

The final ingredient we need to construct manifestly supersymmetric actions is to be able to integrate over superspace.

**Berezin integrals.**

We start by recalling the basics of Berezin integrals. Consider a Grassmann (anticommuting) variable \(\theta\), i.e. \(\theta^2 = 0\), The Berezin integral over this single variable is defined to be a linear functional \[\int d\theta\,(c_1 f_1(\theta) + c_2 f_2(\theta)) = c_1 \int d\theta \, f_1(\theta) + c_2 \int d\theta \, f_2(\theta) ~,\] such that \[\int d\theta \, 1 = 0 ~, \qquad \int d\theta \, \theta = 1 ~.\] These properties uniquely define the Berezin integral. The most general function of \(\theta\) is \(f(\theta) = a + b\theta\) since \(\theta^2 = 0\). Therefore, \[\int d\theta\,f(\theta) = \int d\theta\,(a+b\theta) = b = \frac{d}{d\theta} (a+b\theta) = \frac{d}{d\theta} f(\theta) ~,\] and schematically we have that \(\int d\theta = \frac{d}{d\theta}\). From the form of the general function \(f(\theta)\), it also follows that the Berezin integral of a total derivative vanishes \[\int d\theta\,\frac{d}{d\theta}f(\theta) = 0 ~,\] and that the Berezin integral is translation invariant \[\int d\theta \, f(\theta + \varepsilon) = \int d\theta \, f(\theta) ~.\]

**Integration over \(\mathcal{N}=1\) superspace.**

Generalising to \(\mathcal{N}=1\) superspace we define \[\begin{split} & d^2\theta = \frac{1}{2}d\theta^1 d\theta^2 ~, \qquad d^2\bar\theta = \frac{1}{2}d\bar\theta^2 d\bar\theta^1 ~, \\ & \int d\theta^1d\theta^2\,\theta^2\theta^1 = \int d\bar\theta^2d\bar\theta^1\,\bar\theta^1\bar\theta^2 = 1 ~. \end{split}\] This implies that \[\int d^2\theta \, (\theta\theta) = \int d^2\bar\theta \, (\bar\theta\bar\theta) = \int d^2\theta d^2\bar\theta \, (\theta\theta) (\bar\theta\bar\theta) = 1 ~,\] and schematically we have that \[\int d^2\theta = \frac{1}{4}\epsilon^{\alpha\beta}\partial_\alpha\partial_\beta ~, \qquad \int d^2\bar\theta = -\frac{1}{4}\epsilon^{\dot\alpha\dot\beta}\bar\partial_{\dot\alpha}\bar\partial_{\dot\beta} ~.\]

**Manifestly supersymmetric integrals.**

We can now write down three types of manifestly supersymmetric integral. The first type are integrals of the form \[\begin{equation} \label{eq:type1} \int d^4 x d^2\theta d^2\bar\theta \, Y(x,\theta,\bar\theta) ~, \end{equation}\] where \(Y\) is a general superfield. Since \[\delta_{\varepsilon,\bar\varepsilon} Y = i [\varepsilon Q + \bar\varepsilon \bar Q,Y] = \partial_\alpha\big(-\varepsilon^{\alpha} Y\big) + \bar\partial_{\dot\alpha}\big(-\bar\varepsilon^{\dot\alpha} Y\big) +\partial_\mu\big(- i(\varepsilon\sigma^\mu\bar\theta) Y + i (\theta \sigma^\mu\bar\varepsilon)Y\big) ~,\] is a total superspace derivative, it follows that \[\int d^4 x d^2\theta d^2\bar\theta \, \delta_{\varepsilon,\bar\varepsilon} Y(x,\theta,\bar\theta) = 0~,\] which demonstrates that such integrals are supersymmetric as claimed.

The second type are integrals of the form \[\begin{equation} \label{eq:type2} \int d^4y d^2\theta \, W(y,\theta) = \int d^4xd^2\theta \, W(x,\theta,\bar\theta) ~, \end{equation}\] where \(W\) is a chiral superfield. Note that when we write the integral in terms of the coordinates \((x,\theta,\bar\theta)\) the dependence on \(\bar\theta\) drops out after integrating over \(\theta\). Since \[\delta_{\varepsilon,\bar\varepsilon} W(y,\theta) = i [\varepsilon Q + \bar\varepsilon \bar Q,W(y,\theta)] = \hat\partial_\alpha \big(-\varepsilon^\alpha W(y,\theta)\big) + \hat\partial_{\mu} \big( 2 i (\theta\sigma^\mu \bar\varepsilon)W(y,\theta)\big) ~,\] where \((\hat \partial_\mu,\hat\partial_\alpha,\hat{\bar\partial}_{\dot\alpha})\) denote partial derivatives with respect to \((y,\theta,\bar\theta)\), is a total derivative in chiral superspace, it follows that \[\int d^4y d^2\theta \, \delta_{\varepsilon,\bar\varepsilon} W(y,\theta) = 0 ~,\] demonstrating that such integrals are indeed supersymmetric.

The third type are integrals of the form \[\begin{equation} \label{eq:type3} \int d^4\bar y d^2\bar\theta \, \bar W(\bar y,\bar\theta) = \int d^4xd^2\bar\theta \, \bar W(x,\theta,\bar\theta) ~, \end{equation}\] where \(\bar W\) is an antichiral superfield. An analogous derivation as for the second type shows again that these integrals are supersymmetric.

Note that an integral of the first type can always be written as an integral of the second or third type. To see this we observe that \[\begin{split}
\int d^2\bar\theta \, Y = -\frac{1}{4}\epsilon^{\dot\alpha\dot\beta}\bar\partial_{\dot\alpha}\bar\partial_{\dot\beta} Y & = -\frac{1}{4}\epsilon^{\dot\alpha\dot\beta}\big(\bar D_{\dot\alpha} -i (\theta\sigma^\mu)_{\dot\alpha} \partial_\mu\big)\big(\bar D_{\dot\beta} - i (\theta\sigma^\nu)_{\dot\beta} \partial_\nu\big)Y
\\ & = -\frac{1}{4}\bar D^2Y + \frac{i}{2}\partial_\mu\big(\epsilon^{\dot\alpha\dot\beta}(\theta\sigma^\mu)_{\dot\alpha}\bar D_{\dot\beta}Y\big) + \frac{1}{2} \partial_\mu\partial_\nu\big((\theta\sigma^\mu)_{\dot\alpha}(\theta\sigma^\nu)_{\dot\beta}Y\big) ~,
\end{split}\] hence \[\begin{equation} \label{eq:chirallyexact}
\int d^4x d^2\theta d^2\bar\theta \, Y = -\frac{1}{4}\int d^4x d^2\theta\,\bar D^2Y ~, \end{equation}\] where we have assumed that integrals of total derivatives vanish. Changing coordinates from \((x,\theta,\bar\theta)\) to \((y,\theta,\bar\theta)\), in terms of which \(\bar D_{\dot\alpha} = \hat{\bar\partial}_{\dot\alpha}\), we have \(\bar D_{\dot\alpha}\bar D^2Y = \hat{\bar\partial}_{\dot\alpha} \hat{\bar\partial}^2 Y = 0\). ^{18} Therefore, \(\bar D^2Y\) is a chiral superfield since it only depends on \((y,\theta)\), and the integral \(\eqref{eq:chirallyexact}\) is indeed of the second type. An analogous derivation holds to show that an integral of the first type can be written as an integral of the third type. We say that a chiral superfield that can be written as \(\bar D^2Y\) is chirally exact. Not all chiral superfields are chirally exact and an integral of the second type cannot be written as a (local) integral of the first type unless \(W\) is chirally exact.

Substituting in the expressions for a general superfield \(Y(x,\theta,\bar\theta)\) and a chiral superfield \(\Phi(y,\theta)\) we have \[\int d^4 x d^2\theta d^2\bar\theta \, Y(x,\theta,\bar\theta) = \int d^4 x \, D(x) ~, \qquad \int d^4 y d^2\theta \, \Phi(y,\theta) = - \int d^4x \, F(x) ~.\] Therefore, we call integrals over chiral superspace, which cannot be written as (local) integrals over all superspace, F-terms. These are integrals of the second type that cannot be written as integrals of the first type, i.e. the integrand is chiral, but not chirally exact. Similarly, we call integrals over antichiral superspace, which cannot be written as (local) integrals over all superspace, \(\bar{\text{F}}\)-terms. Local integrals over all superspace are called D-terms.

Finally, supersymmetric actions are constructed from manifestly supersymmetric integrals that

are real, e.g. a real D-term or an F-term plus its conjugate;

are Lorentz scalars, i.e. there are no free indices;

have dimension 0 (including the measure).

In units where \(\hbar = c =1\), we denote the dimension by square brackets, i.e. \([\text{mass}] = [\text{energy}] = [\text{momentum}] = - [\text{time}] = - [\text{length}] = 1\). Therefore, \[\phantom{}[P_\mu] = 1 \qquad \Rightarrow \qquad [x^\mu] = -1 \qquad \Rightarrow \qquad \Big[\int d^4 x\Big] = -4 ~.\] The anticommutation relation \(\{Q_\alpha,\bar Q_{\dot\alpha}\} = 2 (\sigma^{\mu})_{\alpha\dot\alpha} P_\mu\) then implies that \[\phantom{}[Q_\alpha] = [\bar Q_{\dot\alpha}] = \frac{1}{2} \qquad \Rightarrow \qquad [\theta^\alpha] = [\bar\theta^{\dot\alpha}] = - \frac{1}{2} \qquad \Rightarrow \qquad \Big[\int d^2 \theta\Big] = \Big[\int d^2\bar\theta\Big] = 1 ~.\] Note that the dimension of the measure \(\int d\theta\) in a Berezin integral has the opposite sign to the dimension of \(\theta\) since \(\int d\theta\,\theta = 1\). It follows that for the integrals \(\eqref{eq:type1}\), \(\eqref{eq:type2}\) and \(\eqref{eq:type3}\) to be dimensionless we need \([Y] = 2\) and \([W] = [\bar W] = 3\).

We are now finally in a position to construct our first supersymmetric quantum field theories. In this section we will study SQFTs of chiral multiplets. We will will focus on two classes of such theories. These are the Wess-Zumino models, which are renormalisable QFTs, and non-linear sigma models, which are non-renormalisable, but are still of interest as low-energy effective field theories.

The field content of our theories will be \(n\) chiral superfields \(\Phi^i\), where the index \(i=1,\dots,n\). Recall that this means that \(\bar D_{\dot\alpha} \Phi^i = 0\) and we have the component field expansion \[\Phi^i(y,\theta) = \phi^i(y) + \sqrt{2} \theta \psi^i(y) - \theta\theta F^i(y) ~, \qquad y^\mu = x^\mu + i \theta\sigma^\mu\bar\theta ~.\] The conjugate antichiral superfields (\(\bar\Phi^{\bar\imath} = (\Phi^i)^\dagger\)), where \(D_\alpha \bar\Phi^{\bar\imath} = 0\), have the component field expansion \[\bar\Phi^{\bar\imath}(\bar y,\bar \theta) = \bar\phi^{\bar\imath}(\bar y) + \sqrt{2} \bar\theta \bar\psi^{\bar\imath}(\bar y) - \bar\theta\bar\theta \bar F^{\bar \imath}(\bar y) ~, \qquad \bar y^\mu = \bar x^\mu - i \theta\sigma^\mu\bar\theta ~,\] where \(\bar\phi^{\bar\imath} = (\phi^i)^\dagger\), \(\bar\psi^{\bar\imath} = (\psi^i)^\dagger\) and \(\bar F^{\bar\imath} = (F^i)^\dagger\).

The \(\mathcal{N} = 1\) super-Poincaré algebra admits a \(\mathrm{U}(1)\) R-symmetry (\(\mathrm{U}(1)_{_R}\)), which acts on the supercharges as \[\phantom{}[R,Q_{\alpha}] = - Q_{\alpha} ~, \qquad [R,\bar Q_{\dot\alpha}] = \bar Q_{\dot\alpha} ~.\] Denoting the R-charge, i.e. the charge of an object under \(\mathrm{U}(1)_{_R}\), by \([\cdot]_{_R}\), these relations tell us that \[\begin{aligned} \phantom{}[Q_\alpha]_{_R} & = -1 ~, & \qquad [\bar Q_{\dot\alpha}]_{_R} & = 1 ~. \\ \phantom{}[\theta^\alpha]_{_R} & = 1 ~, & \qquad [\bar\theta^{\dot\alpha}]_{_R} & = -1 ~. \end{aligned}\] This follows from the differential operator realisation of the supercharges, \(Q_\alpha = -i\big(\partial_\alpha + \dots\big)\) and \(\bar Q_{\dot\alpha} = i \big(\bar\partial_{\dot\alpha} + \dots\big)\). Therefore, under \(\mathrm{U}(1)_{_R}\) transformations, which we will abbreviate to \(\mathrm{U}(1)_{_R}\) transformations, with parameter \(\alpha\) we have \[\begin{aligned} \theta & \to e^{i\alpha}\theta ~, & \qquad \Phi & \to e^{i\alpha [\Phi]_{_R}}\Phi \qquad \Rightarrow \qquad \begin{cases} \phi \to e^{i[\Phi]_{_R}\alpha}\phi \\ \psi \to e^{i([\Phi]_{_R}-1)\alpha}\psi \\ F \to e^{i([\Phi]_{_R}-2)\alpha}F \end{cases} ~, \end{aligned}\] and their conjugates transform in the opposite way, i.e. with the opposite R-charge.

Our theories may also have global flavour symmetries, which commute with the supercharges \[\phantom{}[F_A,Q_\alpha] = [F_A,\bar Q_{\dot\alpha}] = 0 ~.\] Therefore, the superspace coordinates do not transform under these symmetries and all component fields of a superfield have the same charge.

The most general supersymmetric action of chiral superfields (with at most two derivatives) takes the form \[\begin{equation} \label{eq:globalsym} \mathcal{S}= \int d^4 x d^2\theta d^2\bar\theta \, K(\Phi,\bar\Phi) + \int d^4 x d^2\theta \, W(\Phi) + \int d^4x d^2\bar\theta \, \bar W(\bar\Phi) ~. \end{equation}\] The first term is a D-term, the second an F-term and the third an \(\bar{\text{F}}\)-term.

\(K(\Phi,\bar\Phi)\) in the first term is the Kähler potential and it gives rise to kinetic terms for the component fields. It should have dimension 2 and be a real function of \(\Phi^i\) and \(\bar\Phi^{\bar\imath}\), i.e. it is a composite real scalar field such that \((K(\Phi,\bar\Phi))^\dagger = K(\Phi,\bar\Phi)\). Furthermore, the action is invariant under the Kähler transformation \[K(\Phi,\bar\Phi) \to K(\Phi,\bar\Phi) + \Lambda(\Phi) + \bar\Lambda(\bar\Phi) ~,\] where \(\Lambda(\Phi)\) is a holomorphic function and \(\bar\Lambda(\bar\Phi)\) is an antiholomorphic function. This follows since \(\Lambda(\Phi)\) and \(\bar\Lambda(\bar\Phi)\) are chiral and antichiral superfields, hence \[\int d^4x d^2\theta d^2\bar\theta \, \Lambda(\Phi) = \int d^4x d^2\theta d^2\bar\theta \, \bar\Lambda(\bar\Phi) = 0 ~.\] The physical quantity that is invariant under Kähler transformations is the Kähler metric \[g_{i\bar\jmath}(\Phi,\bar\Phi) = \partial_i \partial_{\bar\jmath} K(\Phi,\bar\Phi) ~,\] where \(\partial_i = \frac{\partial}{\partial\Phi^i}\) and \(\partial_{\bar\imath} = \frac{\partial}{\partial\bar\Phi^{\bar\imath}}\). The Kähler metric is a metric on the complex manifold parametrised by the complex coordinates \(\Phi^i\).

\(W(\Phi)\) in the second term is the superpotential. It should have dimension 3 and be a holomorphic function of \(\Phi^i\), i.e. it is a composite chiral superfield since \(\bar D_{\dot\alpha} W(\Phi) = (\bar D_{\dot\alpha}\Phi^i)\partial_i W(\Phi) = 0\). \(\bar W(\bar\Phi)\) in the third term is the conjugate of \(W(\Phi)\).

The action \(\eqref{eq:globalsym}\) can have various global symmetries:

If \([K]_{_R} = 0\) and \([W]_{_R} = 2\) then \(\mathrm{U}(1)_{_R}\) is a symmetry of the action. This follows from \([\int d^2\theta]_{_R} = -2\) and \([\int d^2\bar\theta]_{_R} = 2\).

There may also be flavour symmetries that leave both \(K\) and \(W\) invariant.

Of course, parameters in the action should not transform under the various global symmetries. However, it may be the case that the global symmetries are explicitly broken, by which we mean that they can formally be restored by assigning non-zero charges to parameters in the action. As an example consider \[K(\Phi,\bar\Phi) = \bar\Phi\Phi ~, \qquad W(\Phi) = \frac{m}{2}\Phi^2 + \frac{\lambda}{3}\Phi^3 ~.\] The resulting action is invariant under \(\mathrm{U}(1)_{_R}\) with \([\Phi]_{_R} = 1\) only if \(\lambda = 0\) since \([m]_{_R} = 0\). The symmetry is broken if \(\lambda\neq0\); however, it can be formally restored by setting \([\lambda]_{_R} = -1\), which implies \([W]_{_R} = 2\) as required.

It is natural to ask why we should assign charges to parameters. This goes back to an old idea in QFT, which is to view parameters that explicitly break a symmetry as background values of non-dynamical external fields that are charged under the broken symmetry. These fields are called spurions. If such a field were dynamical at higher energies, then it would break the symmetry spontaneously. A low-energy observer would not be able to tell whether or not the parameter is actually the vacuum expectation value of such a field. Therefore, we may argue that we should treat parameters (understood as non-dynamical fields) and dynamical fields on an equal footing.

Promoting parameters to background values of non-dynamical fields then restores the symmetry and leads to selection rules that constrain the theory. In SQFT we promote parameters in the superpotential to non-dynamical chiral superfields. This idea is at the root of one of the most powerful results in SQFT: the non-renormalisation theorem for the superpotential.

**Action.**

The Wess-Zumino models are renormalisable theories of chiral superfields. We assign the canonical dimensions to the component fields \[\phantom{}[\phi^i] = 1 ~, \qquad \phantom{}[\psi^i] = \frac{3}{2} ~, \qquad \phantom{}[F^i] = 2 \qquad \Rightarrow \qquad \phantom{}[\Phi^i] = 1 ~,\] and assume that there are no irrelevant couplings. Requiring \([K] = 2\) restricts \(K\) to be quadratic in \(\Phi\) and \(\bar\Phi\). Terms such as \(a_{ij}\Phi^i\Phi^j\) and \(\bar a_{\bar\imath\bar\jmath} \bar\Phi^{\bar\imath}\bar\Phi^{\bar\jmath}\) can be removed by a Kähler transformation. We can then choose a basis of the chiral superfields, i.e. \(\Phi^i \to r^i{}_j \Phi_j\), such that the term containing one \(\Phi\) and one \(\bar\Phi\) is \[\begin{equation} \label{eq:cankahler} K(\Phi,\bar\Phi) = \delta_{i\bar\imath}\bar\Phi^{\bar\imath}\Phi^i = \bar\Phi_i \Phi^i ~. \end{equation}\] This is the canonical Kähler potential. Requiring \([W] = 3\) restricts \(W\) to be a cubic superpotential of the form \[\begin{equation} \label{eq:cubic} W(\Phi) = \frac{1}{2}m_{ij}\Phi^i\Phi^j + \frac{1}{3}\lambda_{ijk}\Phi^i\Phi^j\Phi^k ~, \end{equation}\] where \([m_{ij}]=1\) and \([\lambda_{ijk}]=0\). \(m_{ij}\) is the complex mass matrix and \(\lambda_{ijk}\) are dimensionless coupling constants. Note that only the totally symmetric parts of \(m_{ij}\) and \(\lambda_{ijk}\) contribute to the superpotential.

Let us now express the action \(\eqref{eq:globalsym}\) in terms of component fields for the Kähler potential \(\eqref{eq:cankahler}\) and the cubic superpotential \(\eqref{eq:cubic}\). We start by writing a general superpotential in terms of component fields \[\begin{split} W(\Phi) & = W\big(\phi(y) + \sqrt{2}\theta\psi(y) - \theta\theta F(y)\big) \\ & = W(\phi(y)) + \sqrt{2}\partial_i W(\phi(y))\theta \psi^i(y) - \theta\theta F_W(y) ~, \end{split}\] where \[F_W = \partial_i W(\phi) F^i + \frac{1}{2}\partial_i\partial_jW(\phi)\psi^i\psi^j ~.\] Therefore, \[\int d^4xd^2\theta\,W(\Phi) = - \int d^4x \, \big(\partial_i W(\phi) F^i + \frac{1}{2}\partial_i\partial_jW(\phi)\psi^i\psi^j\big) ~.\] Specialising to the cubic superpotential \(\eqref{eq:cubic}\) we find \[\partial_i W (\Phi) = m_{ij}\Phi^j + \lambda_{ijk} \Phi^j\Phi^k ~, \qquad \partial_i \partial_j W(\Phi) = m_{ij} + 2\lambda_{ijk} \Phi^k ~,\] hence \[F_W = m_{ij}F^i \phi^j + \lambda_{ijk}F^i \phi^j\phi^k + \frac{1}{2}m_{ij}\psi^i\psi^j + \lambda_{ijk}\psi^i\psi^j\phi^k ~.\]

Now turning to the Kähler potential, we have \[\int d^4 x d^2\theta d^2\bar\theta \, K(\Phi,\bar\Phi) = \int d^4x \, D_K(x) ~,\] where \(D_K = K\big|_{\theta\theta\bar\theta\bar\theta}\), i.e. the component of \(K\) proportional to \((\theta\theta)(\bar\theta\bar\theta)\). Therefore, for the Kähler potential \(\eqref{eq:cankahler}\) we take the component field expansion of the chiral superfield \(\Phi\), multiply it by the component field expansion of the antichiral superfield \(\bar\Phi\), and extract the \((\theta\theta)(\bar\theta\bar\theta)\) term. Doing so we find \[D_K = \partial_\mu\bar\phi_i \partial^\mu\phi^i - i\bar\psi_i \bar\sigma^\mu \partial_\mu \psi^i + \bar F_i F^i + \text{total derivative} ~,\] where \(\bar\phi_i = (\phi^i)^\dagger\), \(\bar\psi_i = (\psi^i)^\dagger\) and \(\bar F_i = (F^i)^\dagger\).

Compute the expansion in component fields of the superfield \(W(\Phi)\), where \(\Phi\) denotes a set of superfields \(\Phi^i\), and show that the coefficient of \(-\theta\theta\) is \[F_W = \partial_i W(\phi) F^i + \frac{1}{2}\partial_i\partial_jW(\phi)\psi^i\psi^j ~,\] where \(\partial_i = \frac{\partial}{\partial \Phi^i}\).

Show that the component of the canonical Kähler potential \(\bar\Phi_i\Phi^i\) proportional to \((\theta\theta)(\bar\theta\bar\theta)\) is \[D_K = \partial_\mu\bar\phi_i \partial^\mu\phi^i - i\bar\psi_i \bar\sigma^\mu \partial_\mu \psi^i + \bar F_i F^i + \text{total derivative} ~.\] Determine the total derivative.

Expanding \(W(\Phi) = W\big(\phi(y) + \sqrt{2}\theta\psi(y) - \theta\theta F(y)\big)\) around \(\Phi = \phi(y)\) we find \[\begin{split} W(\Phi) & = W(\phi(y)) + \big(\sqrt{2}\theta\psi^i(y) - \theta\theta F^i(y)\big) \partial_i W(\phi(y)) \\ & \quad +\frac{1}{2}\big(\sqrt{2}\theta\psi^i(y) - \theta\theta F^i(y)\big)\big(\sqrt{2}\theta\psi^j(y) - \theta\theta F^j(y)\big)\partial_i\partial_jW(\phi(y)) \\ & = W(\phi(y)) + \big(\sqrt{2}\theta\psi^i(y) - \theta\theta F^i(y)\big) \partial_i W(\phi(y)) + (\theta\psi^i(y))(\theta\psi^j(y))\partial_i\partial_jW(\phi(y)) \\ & = W(\phi(y)) + \sqrt{2}\theta\psi^i(y) \partial_i W(\phi(y)) - \theta\theta F^i(y) \partial_i W(\phi(y)) -\frac{1}{2} (\theta\theta) (\psi^i(y)\psi^j(y))\partial_i\partial_jW(\phi(y)) ~. \end{split}\] Therefore, the coefficient of \(-\theta\theta\) is \[F_W = \partial_i W(\phi) F^i + \frac{1}{2}\partial_i\partial_jW(\phi)\psi^i\psi^j ~,\] as claimed.

The canonical Kähler potential is \[\bar\Phi_i\Phi^i = \big(\bar\phi_i(\bar y) + \sqrt{2}\bar\theta\bar\psi_i(\bar y) - \bar\theta\bar\theta\bar F_i(\bar y)\big)\big(\phi^i(y) + \sqrt{2}\theta\psi^i(y) - \theta\theta F^i(y)\big) ~.\] We now recall that \(y^\mu = x^\mu + i \theta\sigma^\mu\bar\theta\) and \(\bar y^\mu = x^\mu - i \theta\sigma^\mu\bar\theta\). Therefore, expanding around \(y^\mu = x^\mu\) and \(\bar y^\mu = x^\mu\), and keeping only those terms proportional to \((\theta\theta)(\bar\theta\bar\theta)\), we find \[\begin{split} \bar\Phi_i\Phi^i & = - \frac{1}{2}(\theta\sigma^\mu\bar\theta)(\theta\sigma^\nu\bar\theta) \partial_\mu\partial_\nu \bar\phi_i \phi^i - \frac{1}{2}\bar\phi_i (\theta\sigma^\mu\bar\theta)(\theta\sigma^\nu\bar\theta) \partial_\mu\partial_\nu \phi^i +(\theta\sigma^\mu\bar\theta)\partial_\mu\bar\phi_i(\theta\sigma^\nu\bar\theta)\partial^\mu\phi^i \\ & \quad -2i(\theta\sigma^\mu\bar\theta) (\bar\theta \partial_\mu \bar\psi_i)(\theta\psi^i) +2i(\bar\theta\bar\psi_i)(\theta\sigma^\mu\bar\theta) (\theta\partial_\mu\psi^i) + (\bar\theta\bar\theta)(\theta\theta)\bar F_i F^i + \dots \\ & = - \frac{1}{4} (\theta\theta)(\bar\theta\bar\theta) \partial_\mu\partial^\mu \bar\phi_i \phi^i - \frac{1}{4}(\theta\theta)(\bar\theta\bar\theta)\bar\phi_i \partial_\mu\partial^\mu \phi^i +\frac{1}{2}(\theta\theta)(\bar\theta\bar\theta)\partial_\mu\bar\phi_i\partial^\mu\phi^i \\ & \quad +i(\theta\theta)(\psi^i\sigma^\mu\bar\theta) (\bar\theta \partial_\mu \bar\psi_i) -i(\theta\theta) (\bar\theta\bar\psi_i)(\partial_\mu\psi^i\sigma^\mu\bar\theta) + (\theta\theta) (\bar\theta\bar\theta)\bar F_i F^i + \dots \\ & = - \frac{1}{4} (\theta\theta)(\bar\theta\bar\theta) \partial_\mu\partial^\mu \bar\phi_i \phi^i - \frac{1}{4}(\theta\theta)(\bar\theta\bar\theta)\bar\phi_i \partial_\mu\partial^\mu \phi^i +\frac{1}{2}(\theta\theta)(\bar\theta\bar\theta)\partial_\mu\bar\phi_i\partial^\mu\phi^i \\ & \quad +\frac{i}{2}(\theta\theta)(\bar\theta\bar\theta)( \partial_\mu \bar\psi_i\bar\sigma^\mu\psi^i) -\frac{i}{2}(\theta\theta)(\bar\theta\bar\theta)(\bar\psi_i\bar\sigma^\mu\partial_\mu\psi^i) + (\theta\theta) (\bar\theta\bar\theta)\bar F_i F^i + \dots \\ & = (\theta\theta) (\bar\theta\bar\theta) \Big( -\frac{1}{4}\partial_\mu\big(\partial^\mu \bar\phi_i \phi^i + \bar\phi_i\partial^\mu\phi^i -2i\bar\psi_i\bar\sigma^\mu\psi^i\big) +\partial_\mu\bar\phi_i \partial^\mu\phi^i - i\bar\psi_i \bar\sigma^\mu \partial_\mu \psi^i + \bar F_i F^i \Big) + \dots ~, \end{split}\] where all fields are evaluated at \(x^\mu\). Therefore, as claimed, we have \[D_K = \partial_\mu\bar\phi_i \partial^\mu\phi^i - i\bar\psi_i \bar\sigma^\mu \partial_\mu \psi^i + \bar F_i F^i + \text{total derivative} ~,\] where the total derivative is given by \[-\frac{1}{4}\partial_\mu\big(\partial^\mu \bar\phi_i \phi^i + \bar\phi_i\partial^\mu\phi^i -2i\bar\psi_i\bar\sigma^\mu\psi^i\big) ~.\]

Therefore, we have \[\begin{split} \mathcal{S}& = \int d^4 x d^2\theta d^2\bar\theta \, \bar\Phi_i\Phi^i + \int d^4 x d^2\theta \, W(\Phi) + \int d^4x d^2\bar\theta \, \bar W(\bar\Phi) \\ & = \int d^4x \, \big(\partial_\mu\bar\phi_i \partial^\mu\phi^i - i\bar\psi_i \bar\sigma^\mu \partial_\mu \psi^i + \bar F_i F^i -\partial_i W(\phi) F^i -\bar\partial^i \bar W(\bar\phi) \bar F_i \\ & \qquad\qquad -\frac{1}{2}\partial_i\partial_jW(\phi)\psi^i\psi^j -\frac{1}{2}\bar\partial^i\bar\partial^j\bar W(\bar\phi)\bar\psi_i\bar\psi_j \big) ~, \end{split}\] where we have left the superpotential unspecified. The first two terms are the kinetic terms for the complex scalars \(\phi^i\) and the Majorana fermions \((\psi^i,\bar\psi_i)\). Note that \(- i\bar\psi_i \bar\sigma^\mu \partial_\mu \psi^i = - i\psi^i\sigma^\mu \partial_\mu \bar\psi_i\) up to a total derivative. \(F^i\) and \(\bar F_i\) only appear quadratically and without kinetic terms. This means that they can be integrated out exactly in the path integral. The integral over \(F^i\) and \(\bar F_i\) is Gaussian and replaces them by the solution of their equations of motion \[\begin{equation} \label{eq:feom} F^i = \bar\partial^i \bar W(\bar\phi) ~, \qquad \bar F_i = \partial_i W(\phi) ~. \end{equation}\] This leads to the following potential for the scalar fields \[\begin{equation} \label{eq:scalarpotential} V(\phi,\bar\phi) = \bar\partial^i \bar W(\bar\phi) \partial_i W(\phi)~. \end{equation}\] The resulting action for the component fields is \[\begin{equation} \label{eq:finalaction} \mathcal{S}= \int d^4x \, \big(\partial_\mu\bar\phi_i \partial^\mu\phi^i - i\bar\psi_i \bar\sigma^\mu \partial_\mu \psi^i - \bar\partial^i \bar W(\bar\phi) \partial_i W(\phi) -\frac{1}{2}\partial_i\partial_jW(\phi)\psi^i\psi^j -\frac{1}{2}\bar\partial^i\bar\partial^j\bar W(\bar\phi)\bar\psi_i\bar\psi_j \big) ~. \end{equation}\]

For the cubic superpotential \(\eqref{eq:cubic}\), the scalar potential \(\eqref{eq:scalarpotential}\) includes mass terms for the scalars together with cubic and quartic couplings \[V(\phi,\bar\phi) = \bar m^{ij} m_{ik} \bar\phi_j \phi^k + \bar m^{ij} \lambda_{ikl} \bar\phi_j \phi^k\phi^l +\bar\lambda^{ijk} m_{il} \bar\phi_j\bar\phi_k\phi^l + \bar\lambda^{ijk}\lambda_{ilm} \bar\phi_j\bar\phi_k\phi^l\phi^m ~,\] while the final two terms of the action \(\eqref{eq:finalaction}\) lead to mass terms for the fermions and Yukawa couplings between the scalars and fermions \[\int d^4 x \,\big(-\frac{1}{2}m_{ij} \psi^i\psi^j - \frac{1}{2}\bar m^{ij} \bar\psi_i\bar\psi_j -\lambda_{ijk} \psi^i\psi^j\phi^k - \bar\lambda^{ijk} \bar\psi_i\bar\psi_j\bar\phi_k\big) ~.\]

Each chiral superfield \(\Phi\) together with its conjugate \(\bar\Phi\) contains four bosonic (the two complex scalars \(\phi\) and \(F\)) and four fermionic (the Majorana fermion \((\psi,\bar\psi)\)) real off-shell degrees of freedom. After integrating out the auxiliary fields, the action \(\eqref{eq:finalaction}\) describes one complex scalar and one Majorana fermion for each chiral superfield, the on-shell degrees of freedom of which give a \(\mathcal{N}=1\) massive chiral multiplet.

**Supersymmetric vacua.**

Vacua in QFT should be invariant under Poincaré transformations. This means that the vacuum expectation values of the scalar fields, which minimise the scalar potential as usual, should be constant, and the vacuum expectation values of the Majorana fermions should vanish. The vacuum expectation values of the auxiliary fields \(F^i\) are determined by their equations of motion \(\eqref{eq:feom}\).

Supersymmetric vacua are those vacua that also satisfy \(\partial_i W(\phi)\big|_{\langle\phi^i\rangle} = 0\), which implies that \(\langle F^i\rangle = 0\) since \(F^i = (\partial_i W(\phi))^\dagger\). Therefore, the set of supersymmetric vacua is given by \[\Big\{\langle\phi^i\rangle = \text{constant},~\langle\psi^i\rangle = \langle F^i \rangle = 0 \Big\} ~.\] Note that \(\partial_i W(\phi)\big|_{\langle\phi^i\rangle} = 0\) implies that the scalar potential \(\eqref{eq:scalarpotential}\) is zero on supersymmetric vacua. This is in agreement with the result that supersymmetric vacua have zero energy derived from the supersymmetry algebra in section 3. It also automatically gives us that the scalar potential is minimised since since the scalar potential is manifestly non-negative. As expected, the vacuum expectation values of supersymmetric vacua are invariant under supersymmetry transformations since \[\begin{split} \delta_{\varepsilon,\bar\varepsilon} \langle\phi^i\rangle & \sim \langle\psi^i\rangle = 0 ~, \\ \delta_{\varepsilon,\bar\varepsilon} \langle\psi^i\rangle & \sim \langle\partial_\mu\phi^i\rangle + \langle F^i \rangle = 0 ~, \\ \delta_{\varepsilon,\bar\varepsilon} \langle F^i\rangle & \sim \langle\partial_\mu\psi^i\rangle = 0 ~. \end{split}\]

Often the scalar potential of supersymmetric theories have flat directions. The corresponding massless fields with no potential are called moduli, and the set of supersymmetric vacua \(\mathcal{M}\) is called the moduli space of supersymmetric vacua.

**Equations of motion.**

One way to compute the equations of motion that follow from the action \[\mathcal{S}= \int d^4 x d^2\theta d^2\bar\theta \, \bar\Phi_i\Phi^i + \int d^4 x d^2\theta \, W(\Phi) + \int d^4x d^2\bar\theta \, \bar W(\bar\Phi) ~,\] is to express it in terms of component fields and vary with respect to \(\phi^i\), \(\psi^i\) and \(F^i\) individually. On the other hand, we could try to vary with respect to the superfield \(\Phi^i\). However, we need to be careful when we do this because the three terms in the action have different measures. Furthermore, while the integrands of the F-term and \(\bar{\text{F}}\)-term are constrained to be chiral and antichiral superfields respectively, the integrand of the D-term is not constrained.

In order to vary with respect to \(\Phi^i\) we need to treat all terms in the action that contain \(\Phi^i\), i.e. the D-term and the F-term, on the same footing. Therefore, we write the D-term as an integral over chiral superspace of a chirally exact superfield \[\int d^4x d^2\theta d^2\bar\theta \, \bar\Phi_i\Phi^i = -\frac{1}{4} \int d^4xd^2\theta \, \bar D^2 (\bar \Phi_i \Phi^i) = -\frac{1}{4} \int d^4xd^2\theta \, (\bar D^2 \bar\Phi_i)\Phi^i ~.\] Now varying with respect to \(\Phi^i\) we find the equation of motion \[\frac{1}{4} \bar D^2 \bar\Phi_i = \partial_i W(\Phi) ~.\] Similarly, writing the D-term as an integral over antichiral superspace of an antichirally exact superfield and varying with respect to \(\bar\Phi_i\) we find the equation of motion \[\frac{1}{4} D^2 \bar\Phi_i = \bar\partial^i \bar W(\bar\Phi) ~.\] Note that evaluating these equations of motion on supersymmetric vacua we find \[\begin{split} \langle \bar D^2 \bar\Phi_i\rangle & = 4 \langle \partial_i W(\Phi)\rangle =4 \langle \partial_i W(\phi)\rangle = 0 ~, \\ \langle D^2 \Phi^i\rangle & = 4 \langle \bar\partial^i \bar W(\bar \Phi)\rangle =4 \langle \bar \partial^i \bar W(\bar \phi)\rangle = 0 ~, \end{split}\] which are examples of the general result that the vacuum expectation value of chirally or antichirally exact superfields is zero on supersymmetric vacua.

Consider the Wess-Zumino model of a single chiral superfield \(\Phi\) with \[K(\Phi,\bar\Phi) = \bar\Phi\Phi ~, \qquad W(\Phi) = \frac{m}{2}\Phi^2 + \frac{\lambda}{3}\Phi^3 ~.\] Argue that this \(W(\Phi)\) is the most general renormalisable superpotential and find the supersymmetric vacua of the theory.

Express the action of the Wess-Zumino model in terms of component fields both before and after integrating out the auxiliary fields. Show that the effective physical mass of \(\phi\) is equal to that of \(\psi\) and is given by \(m_{\text{eff}}(\langle\phi\rangle) = m +2 \lambda\langle\phi\rangle\) when the action is expanded around one of the supersymmetric vacua. How is the quartic coupling in the scalar potential related to the Yukawa coupling? Interpret these two results.

Derive the equations of motion for the component fields \(\phi\), \(\psi\) and \(F\) directly from the action expressed in terms of component fields. Finally, expand the equation of motion for the chiral superfield \(\Phi\) and show it is equivalent to the equations of motion for the component fields.

The dimension of the chiral superfield is \([\Phi] = 1\). The action should have dimension 0, which means that \([W]=3\). To ensure renormalisability we assume that there are no irrelevant couplings. This rules out terms in the superpotential of the form \(\lambda_n \Phi^n\) for \(n>3\) since \([\lambda_n\Phi^n] = [\lambda_n] + n[\Phi] = 3\) implies that \([\lambda_n] = 3-n\). Therefore, \(W(\Phi)\) can at most be a cubic polynomial in \(\Phi\). We can then set the linear term in \(W(\Phi)\) to zero by shifting \(\Phi\), while the constant term can be ignored since \(\int d^2\theta \, 1 = 0\). Therefore, the most general renormalisable superpotential is \[W(\Phi) = \frac{m}{2}\Phi^2 + \frac{\lambda}{3}\Phi^3 ~.\]

The supersymmetric vacua of the theory are the solutions to \(W'(\phi)\big|_{\langle\phi\rangle} = 0\). We have \[W'(\phi) = m\phi + \lambda \phi^2 = \phi(m+\lambda\phi) ~,\] hence the two supersymmetric vacua are \[\langle\phi\rangle = 0 ~, \qquad \langle\phi\rangle = -\frac{m}{\lambda} ~.\]

The action of the Wess-Zumino model in terms of component fields is given by \[\begin{split} \mathcal{S}& = \int d^4x \, \big(\partial_\mu\bar\phi \partial^\mu\phi - i\bar\psi \bar\sigma^\mu \partial_\mu \psi + \bar F F - W'(\phi) F -\bar W'(\bar\phi) \bar F -\frac{1}{2}W''(\phi)\psi\psi -\frac{1}{2}\bar W''(\bar\phi)\bar\psi\bar\psi \big) ~, \end{split}\] where \[\begin{aligned} W'(\phi) & = m\phi + \lambda \phi^2 ~, \qquad & W''(\phi) & = m + 2\lambda \phi ~, \\ \bar W'(\bar \phi) & = \bar m\bar \phi + \bar \lambda \bar \phi^2 ~, \qquad & \bar W''(\bar\phi) & = \bar m + 2\bar \lambda \bar \phi ~. \end{aligned}\] Varying with respect to the auxiliary fields \(F\) and \(\bar F\) we find \[\bar F = W'(\phi) = m\phi + \lambda \phi^2 ~,\qquad F = \bar W'(\bar \phi) = \bar m\bar \phi + \bar \lambda \bar \phi^2 ~.\] Substituting back into the action we find that the action of the Wess-Zumino model in terms of component fields after integrating out the auxiliary fields is given by \[\begin{split} \mathcal{S}& = \int d^4x \, \big(\partial_\mu\bar\phi \partial^\mu\phi - i\bar\psi \bar\sigma^\mu \partial_\mu \psi - (\bar m\bar \phi + \bar \lambda \bar \phi^2 )(m\phi + \lambda \phi^2) -\frac{1}{2}(m + 2\lambda \phi)\psi\psi -\frac{1}{2} (\bar m + 2\bar \lambda \bar \phi )\bar\psi\bar\psi \big) ~. \end{split}\]

To compute the effective masses we expand the action around \(\phi = \langle\phi\rangle\) and \(\psi = 0\) to quadratic order in fluctuations, which we denote by \(\varphi\) and \(\vartheta\), we find \[\begin{split} \mathcal{S}& = \int d^4x \, \big(-(\bar m\langle\bar \phi\rangle + \bar \lambda\langle\bar \phi\rangle^2)(m\langle\phi\rangle + \lambda\langle \phi\rangle^2) \\ & \qquad\qquad -(\bar m\langle\bar \phi\rangle + \bar \lambda\langle\bar \phi\rangle^2)(m +2\lambda)\varphi -(m\langle\phi\rangle + \lambda\langle \phi\rangle^2)(\bar m+2\bar\lambda)\bar\varphi \\ & \qquad \qquad + \partial_\mu\bar{\varphi} \partial^\mu\varphi - (\bar m\langle\bar \phi\rangle + \bar \lambda\langle\bar \phi\rangle^2) \lambda \varphi^2 - (m\langle\phi\rangle + \lambda\langle \phi\rangle^2)\bar\lambda \bar{\varphi}^2 \\ & \qquad\qquad - (\bar m + 2 \bar\lambda\langle\bar \phi\rangle)(m + 2\lambda\langle \phi\rangle)\bar\varphi \varphi \\ & \qquad\qquad - i\bar{\vartheta} \bar\sigma^\mu \partial_\mu \vartheta -\frac{1}{2}(m + 2\lambda \langle\phi\rangle)\vartheta\vartheta -\frac{1}{2} (\bar m + 2\bar \lambda\langle \bar \phi\rangle )\bar{\vartheta}\bar{\vartheta} \big) + \dots ~. \end{split}\] The supersymmetric vacua are \(\langle \phi \rangle = 0\) or \(\langle \phi \rangle = -\frac{m}{\lambda}\), for which the constant term and the terms proportional to \(\varphi\), \(\bar\varphi\), \(\varphi^2\) and \(\bar\varphi^2\) vanish. Therefore, the action becomes \[\begin{split} \mathcal{S}& = \int d^4x \, \big(\partial_\mu\bar{\varphi} \partial^\mu\varphi - (\bar m + 2 \bar\lambda\langle\bar \phi\rangle)(m + 2\lambda\langle \phi\rangle)\bar\varphi \varphi \\ &\qquad\qquad - i\bar{\vartheta} \bar\sigma^\mu \partial_\mu \vartheta -\frac{1}{2}(m + 2\lambda \langle\phi\rangle)\vartheta\vartheta -\frac{1}{2} (\bar m + 2\bar \lambda\langle \bar \phi\rangle )\bar{\vartheta}\bar{\vartheta} \big) + \dots ~, \end{split}\] and we can read off the effective mass of \(\phi\) and \(\psi\) \[m^{\phi}_{\text{eff}} = m^{\psi}_{\text{eff}} = m_{\text{eff}} = |m + 2\lambda\langle \phi\rangle| ~.\] Note that the effective mass is nothing but \(m_{\text{eff}} = |W''(\phi)|\). Expanding to quartic in order in fluctuations we see that the quartic coupling is \(\bar\lambda\lambda\bar\varphi^2\varphi^2\), while the Yukawa couplings are \(\lambda\varphi\vartheta\vartheta\) and \(\bar\lambda\bar\varphi\bar\vartheta\bar\theta\). Therefore, schematically we have that \((\text{quartic}) = (\text{Yukawa})^2\). This is a consequence of supersymmetry and, together with the equal effective masses, leads to cancellations between quantum corrections due to bosons and those due to fermions around supersymmetric vacua.

The equations of motion for the component fields are \[\begin{aligned} \phi: \quad & \partial_\mu\partial^\mu \bar\phi + W''(\phi)F + \frac{1}{2}W'''(\phi) \psi\psi = 0 ~, & \qquad \bar\phi: \quad & \partial_\mu\partial^\mu \phi + \bar W''(\bar \phi)\bar F + \frac{1}{2}\bar W'''(\bar \phi) \bar \psi\bar \psi = 0 ~, \\ \psi: \quad & i\sigma^\mu\partial_\mu\bar\psi + W''(\phi)\psi = 0 ~, & \qquad \bar\psi: \quad & i\bar \sigma^\mu\partial_\mu\psi + \bar W''(\bar\phi)\bar\psi = 0 ~, \\ F: \quad & \bar F = W'(\phi) ~, & \bar F: \quad & F = \bar W'(\bar\phi) ~. \end{aligned}\] The equations of motion for the chiral and antichiral superfield are \[\bar D^2 \bar\Phi = 4 W'( \Phi) ~, \qquad D^2 \Phi = 4\bar W'(\bar \Phi) ~.\] We start by writing the antichiral superfield \(\bar\Phi(\bar y,\bar\theta)=\bar\phi(\bar y) + \sqrt{2}\bar\theta\bar\psi(\bar y) - \bar\theta\bar\theta \bar F(\bar y)\) in terms of the coordinates \((y,\theta,\bar\theta)\) \[\begin{split} \bar\Phi(y,\theta,\bar\theta) & = \bar\phi(y) + \sqrt{2}\bar\theta\bar\psi(y) - \bar\theta\bar\theta \bar F(y) \\ & \quad - 2i(\theta\sigma^\mu\bar\theta)\hat{\partial}_\mu\bar\phi(y) - 2\sqrt{2}i(\theta\sigma^\mu\bar\theta)(\bar\theta\hat{\partial}_\mu\bar\psi(y)) - 2(\theta\sigma^\mu\bar\theta)(\theta\sigma^\nu\bar\theta)\hat{\partial}_\mu\hat{\partial}_\nu \bar\phi(y) \\ & = \bar\phi(y) + \sqrt{2}\bar\theta\bar\psi(y) - \bar\theta\bar\theta \bar F(y) \\ & \quad - 2i(\theta\sigma^\mu\bar\theta)\hat{\partial}_\mu\bar\phi(y) + \sqrt{2}i(\bar\theta\bar\theta)(\theta\sigma^\mu\hat{\partial}_\mu\bar\psi(y)) - (\theta\theta)(\bar\theta\bar\theta)\hat{\partial}_\mu\hat{\partial}^\mu \bar\phi(y) ~, \end{split}\] where we have used that \(\bar y^\mu = y^\mu - 2i\theta\sigma^\mu\bar\theta\). In terms of derivatives with respect to \((y,\theta,\bar\theta)\), \(\bar D_{\dot\alpha} = \hat{\bar\partial}_{\dot\alpha}\). Using that \(\hat{\bar\partial}^2(\bar\theta\bar\theta) = -4\) we find \[\bar D^2 \bar\Phi (y,\theta) = 4\big(\bar F(y) - \sqrt{2}i\theta\sigma^\mu\hat{\partial}_\mu\bar\psi(y) + \theta\theta\hat{\partial}_\mu\hat{\partial}^\mu \bar\phi(y)\big) ~.\] Next we observe that \[\begin{split} 4W'(\Phi(y,\theta)) & = 4 W'(\phi(y)+\sqrt{2}\theta \psi(y) - \theta\theta F(y)) \\ & = 4 \big(W'(\phi(y)) + (\sqrt{2}\theta \psi(y) - \theta\theta F(y)) W''(\phi(y)) + (\theta\psi(y))(\theta\psi(y)) W'''(\phi(y))\big) \\ & = 4 \big(W'(\phi(y)) + \sqrt{2}\theta \psi(y)W''(\phi(y)) - \theta\theta (F(y) W''(\phi(y)) +\frac{1}{2} \psi(y)\psi(y) W'''(\phi(y)))\big) ~. \end{split}\] Equating \(\bar D^2 \bar\Phi(y,\theta)\) with \(4W'(\Phi(y,\theta))\), we recover the equations of motion for the component fields \[\bar F = W'(\phi) ~, \qquad i\sigma^\mu\partial_\mu\bar\psi + W''(\phi) \psi = 0 ~, \qquad \partial_\mu\partial^\mu \bar\phi + W''(\phi) F + \frac{1}{2}W'''(\phi) \psi\psi = 0 ~,\] as the coefficients of \(1\), \(\theta\) and \(\theta\theta\) respectively, where we have set \(y^\mu = x^\mu\). A similar derivation holds for the conjugate equation of motion \(D^2\Phi = 4\bar W'(\bar\Phi)\).

Consider the Wess-Zumino model of three chiral superfields \(X\), \(Y\) and \(Z\) with the homogeneous trilinear superpotential \(W(X,Y,Z) = \lambda XYZ\). Write down the scalar potential of the theory. Find the moduli space of supersymmetric vacua.

The scalar potential of the theory is given by \[\begin{split} V(x,y,z,\bar x,\bar y,\bar z) & = \sum_{i = x,y,z} \bar\partial^i \bar W(\bar x,\bar y,\bar z) \partial_i W(x,y,z) \\ & = \bar\lambda\lambda (\bar y y \bar z z + \bar x x \bar z z + \bar x x \bar y y) ~, \end{split}\] where the component field expansions of the chiral superfields \(X\), \(Y\) and \(Z\) are \[X = x + \mathcal{O}(\theta) ~, \qquad Y = y + \mathcal{O}(\theta) ~, \qquad Z = z + \mathcal{O}(\theta) ~.\] The moduli space of supersymmetric vacua is the space of solutions to \(\partial_i W(x,y,z)\big|_{\langle x\rangle,\langle y\rangle,\langle z\rangle} = 0\). This gives us the following three equations \[\langle y \rangle\langle z \rangle = \langle x\rangle\langle z \rangle = \langle x\rangle\langle y \rangle = 0 ~.\] These are solved by setting any two of \(\langle x\rangle\), \(\langle y\rangle\) and \(\langle z\rangle\) equal to zero. There are three ways to do this, each of which gives a branch or component of the moduli space of vacua. The moduli space of supersymmetric vacua is therefore given by \[\begin{split} \mathcal{M} & = \mathcal{M}_x \cup \mathcal{M}_y \cup \mathcal{M}_z ~, \\ \mathcal{M}_x & = \{x \in \mathbb{C}, ~ y=z = 0\} \cong \mathbb{C}~, \\ \mathcal{M}_y & = \{y \in \mathbb{C}, ~ x=z = 0\} \cong \mathbb{C}~, \\ \mathcal{M}_z & = \{z \in \mathbb{C}, ~ x=y = 0\} \cong \mathbb{C}~. \end{split}\] The three branches intersect at the origin.

Theories with a general Kähler potential are typically not renormalisable, but they are still interesting as low-energy supersymmetric effective field theories (EFTs) or interacting superconformal field theories (SCFTs). They are the most general theories of chiral multiplets with at most two derivatives, which follows from the fact that the Kähler potential only depends on \(\Phi^i\) and \(\bar\Phi^i\) and not their derivatives.

Non-linear sigma models usually describe massless fields that have a (continuous) moduli space of vacua. Therefore, the superpotential is typically zero. Nevertheless, one can include it, which is useful when studying interacting SCFTs.

Expanding out the non-linear sigma model action in terms of component fields we find \[\begin{split} \mathcal{S}& = \int d^4 x d^2\theta d^2\bar\theta \, K(\Phi,\bar\Phi) + \int d^4 x d^2\theta \, W(\Phi) + \int d^4x d^2\bar\theta \, \bar W(\bar\Phi) \\ & = \int d^4x \, \big( g_{i{\bar\jmath}}(\phi,\bar\phi)(\partial_\mu\phi^i\partial^\mu\bar\phi^{\bar\jmath} - \frac{i}{2} \bar\psi^{\bar\jmath} \bar \sigma^\mu D_\mu\psi^i - \frac{i}{2} \psi^i \sigma^\mu D_\mu \bar\psi^{\bar\jmath}) - g^{i\bar\jmath}(\phi,\bar\phi)\partial_i W(\phi)\bar\partial_{\bar\jmath}\bar W(\bar\phi) \\ & \qquad\qquad - \frac{1}{2}(\partial_i\partial_j W(\phi) - \Gamma^k_{ij}(\phi,\bar\phi)\partial_k W(\phi)) - \frac{1}{2}(\bar\partial_{\bar\imath}\bar\partial_{\bar\jmath} \bar W(\bar\phi) - \bar\Gamma^{\bar k}_{\bar\imath\bar\jmath}(\phi,\bar\phi)\bar\partial_{\bar k} \bar W(\bar\phi)) \\ & \qquad \qquad + \frac{1}{4} R_{i\bar\jmath k \bar l} (\phi,\bar\phi) \psi^i \psi^k\bar\psi^{\bar\jmath}\bar\psi^{\bar l} \big) ~, \end{split}\] where \(g_{i{\bar\jmath}}(\phi,\bar\phi) = \partial_i\bar\partial_{\bar\jmath} K(\phi,\bar\phi)\) is the Kähler metric and \(g^{i{\bar\jmath}}(\phi,\bar\phi)\) is its inverse. \(\Gamma^k_{ij}(\phi,\bar\phi)\) and \(\bar\Gamma^{\bar k}_{\bar\imath\bar\jmath}(\phi,\bar\phi)\) are the non-zero components of the Levi-Civita connections – the mixed holomorphic and antiholomorphic components vanish – and \(R_{i\bar\jmath k \bar l} (\phi,\bar\phi)\) is the Riemann tensor. \(D_\mu\psi^i = \partial_\mu \psi^i + \Gamma^i_{jk}\partial_\mu\phi^j \psi^k\) and \(D_\mu \bar\psi^{\bar\imath} = \partial_\mu\bar\psi^{\bar\imath} + \bar\Gamma^{\bar \imath}_{\bar\jmath\bar k} \partial_\mu \bar\phi^{\bar\jmath} \bar\psi^{\bar k}\) are Kähler covariant derivatives.

The target space geometry with metric \(g_{i{\bar\jmath}} (\phi,\bar\phi) d\phi^i d\bar\phi^{\bar\jmath} = \partial_i\bar\partial_{\bar\jmath} K (\phi,\bar\phi) d\phi^i d\bar\phi^{\bar\jmath}\) is a Kähler manifold. This is a consequence of \(\mathcal{N}=1\) supersymmetry. A Kähler manifold is a complex hermitian manifold with a non-degenerate closed (1,1)-form, i.e. a form with one holomorphic and one antiholomorphic index \[\omega = \frac{1}{2}\omega_{i\bar\jmath}(\phi,\bar\phi)d\phi^i\wedge d\bar\phi^{\bar\jmath} ~.\] Locally, we have that \(\omega_{i\bar\jmath} = g_{i\bar\jmath}\).

Let us now see an example of the power of supersymmetry in QFT. We will use supersymmetry and holomorphy to show that the superpotential in a theory of chiral multiplets is not renormalised. There is a perturbative proof using supergraphs due to Grisaru, Siegel and Roček, but we will give a version that is fully non-perturbative due to Seiberg.

The Wilsonian effective action is obtained by integrating out those modes with Euclidean momentum \(|p| > \mu\). Given the Wilsonian effective action at scale \(\mu\), the effective action at scale \(\mu-\delta\mu\) is given by integrating out modes with \(\mu-\delta\mu < |p| \leq \mu\). As we integrate out more modes, the values of the couplings and the wave functions may change, leading to their renormalisation. This is known as the Wilsonian renormalisation group.

For an SQFT of chiral multiplets, the leading terms in the Wilsonian effective action should take the general form \[\mathcal{S}_{\text{eff},\mu} = \int d^4 x d^2\theta d^2\bar\theta \, K_{\text{eff},\mu}(\Phi,\bar\Phi) + \int d^4 x d^2\theta \, W_{\text{eff},\mu}(\Phi) + \int d^4x d^2\bar\theta \, \bar W_{\text{eff},\mu}(\bar\Phi) ~.\] The quantum corrections coming from integrating out degrees of freedom may result in \(K_{\text{eff},\mu}\) and \(W_{\text{eff},\mu}\) depending on the renormalisation group scale \(\mu\). In particular, they may be different from \(K_{\text{micro}} = K_{\text{eff},\Lambda_{\text{UV}}}\) and \(W_{\text{micro}} = W_{\text{eff},\Lambda_{\text{UV}}}\), the values of \(K\) and \(W\) at the UV scale \(\Lambda_{\text{UV}}\) at which the theory is defined. If the theory is UV complete, we can take \(\Lambda_{\text{UV}} \to \infty\), and \(K_{\text{micro}}\) and \(W_{\text{micro}}\) are the bare Kähler potential and superpotential.

What we would like to show is that \(W_{\text{eff},\mu} = W_{\text{micro}}\), i.e. the superpotential is not renormalised. This is known as a non-renormalisation theorem. To do this we will use three key ideas to constrain the form of \(W_{\text{eff},\mu}\):

holomorphy in the microscopic coupling constants;

selection rules from broken symmetries, under which the coupling constants may transform;

smoothness of the physics in weak coupling limits.

The first two of these follow from viewing the coupling constants in the superpotential as background chiral superfields, i.e. as spurions. For simplicity we will consider the superpotential \[W_{\text{micro}} = \frac{m}{2}\Phi^2 + \frac{\lambda}{3}\Phi^3 ~.\] Let us now look at the implications of these three key ideas.

Holomorphy implies that \(W_{\text{eff},\mu}\) is a holomorphic function of \(\Phi\), \(m\) and \(\lambda\), i.e. it does not depend on \(\bar\Phi\), \(\bar m\) and \(\bar\lambda\). It may also depend on the renormalisation group scale \(\mu\).

There are two broken symmetries. The first of these is the \(\mathrm{U}(1)_{_R}\) symmetry under which the superpotential and chiral superfield have charges \([W]_{_R} = 2\) and \([\Phi]_{_R} = 1\). Therefore, the coupling constants have charges \([m]_{_R} = 0\) and \([\lambda]_{_R} = -1\). There is also an additional \(\mathrm{U}(1)\) flavour symmetry (\(\mathrm{U}(1)_{_F}\)) whose charge we denote \([\cdot]_{_F}\). Under the \(\mathrm{U}(1)_{_F}\) symmetry the superpotential should not be charged. Therefore, setting \([\Phi]_{_F} = 1\), we have \([m]_{_F} = -2\) and \([\lambda]_{_F} = -3\).

Requiring that \([W_{\text{eff},\mu}]_{_R} = 2\) and \([W_{\text{eff},\mu}]_{_F} = 0\), the effective superpotential is constrained to take the form \[\begin{equation} \label{eq:ansatzeff} W_{\text{eff},\mu} = m\Phi^2 f(t,\mu) ~, \qquad t = \frac{\lambda\Phi}m ~. \end{equation}\] This follows since \(t=\frac{\lambda\Phi}m\) is the only combination of the superfield and the coupling constants that has zero R-charge and zero flavour charge. Therefore, the effective superpotential can be an arbitrary function of \(t=\frac{\lambda\Phi}m\) multiplied by \(m\Phi^2\) to ensure that \(W_{\text{eff},\mu}\) has the correct overall R-charge and flavour charge. We could multiply by any function with the correct charges, e.g. \(\lambda\Phi^3\), but it is easy to see that the resulting ansatz is equivalent to \(\eqref{eq:ansatzeff}\).

We now consider the limit \(m\to 0\) and \(\lambda\to 0\) with arbitrary \(\frac{\lambda}{m}\). In this limit the theory becomes free and we should have that \(W_{\text{eff},\mu} \to W_{\text{micro}}\) at the first subleading order. Therefore, \(f(t,\mu) \to \frac12+\frac{1}{3}t\), which, since \(t\) is arbitrary in this limit, implies that \(f(t,\mu) = \frac12 + \frac{1}{3}t\).

As a result, we find that \[W_{\text{eff},\mu} = \frac{m}{2}\Phi^2 + \frac{\lambda}{3}\Phi^3 = W_{\text{micro}} ~.\] This holds for any renormalisation group scale \(\mu\), hence the superpotential is not renormalised. Note that the couplings and wave functions may be individually renormalised, but the combinations that appear in the superpotential are not renormalised, i.e. \[\Phi \to Z_\Phi \Phi ~, \qquad m \to Z_m m ~, \qquad \lambda \to Z_\lambda \lambda ~,\] such that \[Z_m Z_\Phi^2 = Z_\lambda Z_\Phi^3 = 1 ~.\]

To conclude, a couple of comments are in order. The first is that the Wess-Zumino model is not UV complete since it is not asymptotically free. The second is that we may be concerned that the limit \(m\to 0\) and \(\lambda\to 0\) is not smooth since the resulting free theory has massless modes, which may lead to IR divergences. However, since we are only integrating out modes with \(|p|>\mu\), the Wilsonian effective action does not have such IR divergences.

Argue that the superpotentials \(W(\Phi) = \sum_{n=2}^N \lambda^n \Phi^n\) and \(W(\Phi) = \frac{1}{2} m_{ij} \Phi^i\Phi^j + \frac{1}{3} \lambda_{ijk}\Phi^i\Phi^j\Phi^k\) are not renormalised in an SQFT of chiral multiplets.

We start with the superpotential \[W(\Phi) = \sum_{n=2}^N \lambda^n \Phi^n ~.\] Holomorphy implies that \(W_{\text{eff},\mu}\) is a holomorphic function of \(\Phi\) and \(\lambda_n\). There are two broken symmetries: \(\mathrm{U}(1)_{_R}\) with \([W]_{_R} = 2\) and \([\Phi]_{_R} = 1\), and \(\mathrm{U}(1)_{_F}\) with \([W]_{_F} = 0\) and \([\Phi]_{_F} = 1\). The coupling constants have charges \([\lambda_n]_{_R} = -n+2\) and \([\lambda_n]_{_F} = -n\). Therefore, the effective superpotential is constrained to take the form \[W_{\text{eff},\mu} = \lambda_2\Phi^2 f(t_i,\mu) ~, \qquad t_i = \frac{\lambda_i\Phi^{i-2}}{\lambda_2} ~,\] where \(t_i = \frac{\lambda_i\Phi^{i-2}}{\lambda_2}\), \(i=3,\dots,N\), are a maximal set of independent combinations of the superfield and the coupling constants with zero R-charge and zero flavour charge. Now taking the limit \(\lambda_n \to 0\) with \(t_i\) fixed the theory becomes free and we should have that \(W_{\text{eff},\mu} \to W\). Therefore, \(f(t_i,\mu) \to 1 + \sum_{n=3}^N t_i\), which immediately implies that \(f(t_i,\mu) = 1 + \sum_{n=3}^N t_i\). As a result, we find that \[W_{\text{eff},\mu} = \sum_{n=2}^N \lambda^n \Phi^n = W ~.\]

For the superpotential \[W(\Phi) = \frac{1}{2} m_{ij} \Phi^i\Phi^j + \frac{1}{3} \lambda_{ijk}\Phi^i\Phi^j\Phi^k ~, \qquad i=1,\dots,n ~,\] holomorphy implies that \(W_{\text{eff},\mu}\) is a holomorphic function of \(\Phi^i\), \(m_{ij}\) and \(\lambda_{ijk}\). The broken symmetries are now \(\mathrm{U}(1)_{_R}\) with \([W]_{_R} = 2\), \([\Phi]_{_R} = 1\), \([m_{ij}]_{_R} = 0\) and \([\lambda_{ijk}]_{_R} = -1\), and \(\mathrm{U}(n)_{_F}\), where \(\Phi^i\) transforms in the fundamental representation, and \(m_{ij}\) and \(\lambda_{ijk}\) transform in the double and triple symmetric tensor products of the antifundamental representation respectively. Let us now choose one chiral superfield \(\Phi^{\hat\imath}\). The superpotential is a cubic polynomial in \(\Phi^{\hat\imath}\) with effective couplings that depend on \(m_{ij}\), \(\lambda_{ijk}\) and \(\Phi^{i\neq\hat{\imath}}\). Therefore, the non-renormalisation argument applies. Since this holds for all \(\Phi^i\), it follows that the superpotential \[W(\Phi) = \frac{1}{2} m_{ij} \Phi^i\Phi^j + \frac{1}{3} \lambda_{ijk}\Phi^i\Phi^j\Phi^k ~,\] is not renormalised.

In section 4 we outlined two ways to reduce the number of component fields in a general superfield. One of these was to impose the chiral constraint. SQFTs of the resulting chiral superfields were the subject of section 5. The other way is to impose a reality condition \[\begin{equation} \label{eq:reality} V = V^\dagger ~. \end{equation}\] SQFTs of the resulting real superfields \[\begin{split} V(x,\theta,\bar\theta) &= C(x) + \theta\chi(x) + \bar\theta\bar\chi(x) +\theta\theta M(x) +\bar\theta\bar\theta\bar M(x) + (\theta\sigma^\mu\bar\theta) A_\mu(x) \\ & \quad + i(\theta\theta) \bar\theta\big(\bar\lambda(x) + \frac{1}{2}\bar\sigma^\mu\partial_\mu\chi(x)\big) - i(\bar\theta\bar\theta) \theta\big(\lambda(x) - \frac{1}{2}\sigma^\mu\partial_\mu\bar\chi(x)\big) \\ & \quad +\frac{1}{2}(\theta\theta)(\bar\theta\bar\theta)\big(D(x) - \frac{1}{2}\partial_\mu\partial^\mu C(x)\big) ~, \end{split}\] where \[C^\dagger = C ~, \qquad M^\dagger = \bar M ~, \qquad A^\dagger_\mu = A_\mu ~, \qquad D^\dagger = D ~,\] are the subject of this section. Note that the reality condition \(\eqref{eq:reality}\) is covariant under supersymmetry since \(\varepsilon Q + \bar\varepsilon \bar Q\) is hermitian. The real superfield has 8 real bosonic degrees of freedom: \(C\) and \(D\) are real scalars, \(M\) is a complex scalar and \(A_\mu\) is a real vector. It also has 8 real fermionic degrees of freedom: \(\chi\) and \(\lambda\) are Weyl fermions.

The real superfield contains a real vector \(A_\mu\). The most important QFTs of vector fields are gauge theories. Therefore, our aim is to use the real superfield to construct supersymmetric gauge theories. To do so we need to introduce a supersymmetric version of gauge transformations.

We start with the construction of abelian supersymmetric gauge theories. Consider a real superfield \(V = V^\dagger\) with gauge symmetry \[\begin{equation} \label{eq:susygauge} V \to V + \Lambda + \bar\Lambda ~, \end{equation}\] where \(\Lambda\) is a chiral superfield. We refer to \(V\) as an abelian vector superfield. The gauge transformations of the component fields of \(V\) are \[\begin{aligned} C & \to C + 2\mathop{\mathrm{Re}}\phi ~, \qquad & A_\mu & \to A_\mu - 2\partial_\mu \mathop{\mathrm{Im}}\phi ~, \\ M & \to M - F ~, \qquad & D & \to D ~, \\ \chi & \to \chi + \sqrt{2}\psi ~, \qquad & \lambda & \to \lambda ~, \end{aligned}\] where \(\phi\), \(\psi\) and \(F\) are the component fields of \(\Lambda\) \[\Lambda(y,\theta) = \phi(y) + \sqrt{2}\theta\psi(y) - \theta\theta F(y) ~.\]

By suitably choosing \(\mathop{\mathrm{Re}}\phi\), \(\psi\) and \(F\) we can gauge away \(C\), \(\chi\) and \(M\), i.e. fix a gauge in which they are equal to zero. In this gauge, known as the Wess-Zumino (WZ) gauge, the abelian vector superfield takes the form \[V_{\text{WZ}}(x,\theta,\bar\theta) = (\theta\sigma^\mu\bar\theta) A_\mu(x) + i (\theta\theta)(\bar\theta\bar\lambda(x)) - i (\bar\theta\bar\theta)(\theta\lambda(x)) + \frac{1}{2}(\theta\theta)(\bar\theta\bar\theta) D(x) ~.\] The component fields that remain are the gauge field \(A_\mu\), the gaugino \(\lambda\) and the real auxiliary field \(D\). The residual gauge freedom is \[A_\mu \to A_\mu - 2\partial_\mu \mathop{\mathrm{Im}}\phi ~.\]

Many computations are easier in the WZ gauge because \[V^2_{\text{WZ}}(x,\theta,\bar\theta) = (\theta \sigma^\mu \bar\theta)(\theta\sigma^\nu\bar\theta)A_\mu(x) A_\nu(x) = \frac{1}{2}(\theta\theta)(\bar\theta\bar\theta) A_\mu(x) A^\mu(x) ~,\] where all other terms vanish since \(\theta\) and \(\bar\theta\) are anticommuting. As a result, we have relations such as \[V_{\text{WZ}}^3 = V_{\text{WZ}}^2 D_\alpha V_{\text{WZ}} = V_{\text{WZ}}^2 \bar D_{\dot\alpha} V_{\text{WZ}} = 0 ~,\] since \(V_{\text{WZ}} \sim \theta \bar\theta\), \(D_\alpha V_{\text{WZ}} \sim \bar\theta\) and \(\bar D_{\dot\alpha} V_{\text{WZ}} \sim \theta\). On the other hand, the WZ gauge is not supersymmetric, i.e. supersymmetry transformations take us out of the WZ gauge. This is straightforward to see from the supersymmetry transformations of the component fields of a general superfield derived in section 4. Therefore, if we start in the WZ gauge and perform a supersymmetry transformation, we need a compensating gauge transformation \(V \to V + \Lambda + \bar\Lambda\), with \(\Lambda\) suitably chosen, to go back to the WZ gauge.

The next step is to find a superfield that is invariant under supersymmetric gauge transformations \(\eqref{eq:susygauge}\), which contains the field strength \(F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu\). This is the gaugino superfield \[W_\alpha = - \frac{1}{4} \bar D^2 D_\alpha V ~, \qquad \bar W_{\dot \alpha} = -\frac{1}{4} D^2 \bar D_{\dot\alpha} V ~.\] Note that \(W_\alpha\) and \(\bar W_{\dot \alpha}\) are chirally and antichirally exact. Under supersymmetric gauge transformations we have \[\begin{split} W_\alpha & \to W_\alpha - \frac{1}{4} \bar D^2 D_\alpha \Lambda - \frac{1}{4} \bar D^2 D_\alpha \bar\Lambda = W_\alpha + \frac{1}{4} \bar D^{\dot \alpha} \bar D_{\dot\alpha} D_\alpha \Lambda = W_\alpha + \frac{1}{4}\bar D^{\dot \alpha} \{\bar D_{\dot\alpha}, D_\alpha \} \Lambda \\ &\quad = W_\alpha + \frac{i}{2}\bar D^{\dot \alpha} (\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu \Lambda = W_\alpha + \frac{i}{2}(\sigma^\mu)_{\alpha\dot\alpha} \epsilon^{\dot\alpha\dot\beta} \partial_\mu\bar D_{\dot \beta}\Lambda = W_\alpha ~, \end{split}\] where we have used that \(D_\alpha \bar \Lambda = \bar D_{\dot\alpha} \Lambda = 0\). Since \(W_\alpha\) is chirally exact, it is convenient to write its component field expansion in terms of the chiral superspace coordinates \((y,\theta,\bar\theta)\) \[W_\alpha(y,\theta) = - i \lambda_\alpha (y) + \theta_\alpha D(y) + i(\sigma^{\mu\nu}\theta)_\alpha F_{\mu\nu}(y) + (\theta\theta)(\sigma^\mu\hat\partial_\mu\bar\lambda(y))_\alpha ~,\] where \(F_{\mu\nu}(y) = \hat\partial_\mu A_\nu(y) - \hat\partial_\nu A_\mu(y)\) is the abelian field strength and we recall that \[(\sigma^{\mu\nu})_{\alpha}{}^\beta = \frac{1}{4}(\sigma^\mu\bar\sigma^\nu - \sigma^\nu\bar\sigma^\mu)_{\alpha}{}^\beta ~, \qquad (\bar\sigma^{\mu\nu})^{\dot\alpha}{}_{\dot\beta} = \frac{1}{4}(\bar\sigma^\mu\sigma^\nu - \bar\sigma^\nu\sigma^\mu)^{\dot\alpha}{}_{\dot\beta} ~.\] \(W_\alpha\) is known as the gaugino superfield because its lowest component is the gaugino \(\lambda\). Indeed, it is clear that \(W_\alpha\) is fermionic, i.e. it is anticommuting and transforms as a Lorentz spinor. It is also sometimes called the super field strength since it is the supersymmetric completion of the field strength \(F_{\mu\nu}\).

Starting from an abelian vector superfield \(V\) and its supersymmetric gauge transformation \(V \to V + \Lambda + \bar\Lambda\), where \(\Lambda\) is a chiral superfield, determine the gauge transformations of the component fields of \(V\).

Fixing the Wess-Zumino (WZ) gauge \(V = V_{\text{WZ}}\), compute the supersymmetry transformation of \(V_{\text{WZ}}\) and determine the compensating gauge transformation that is needed to restore the WZ gauge.

Derive the component field expansion of the gaugino superfield \(W_\alpha = -\frac{1}{4} \bar D^2 D_\alpha V\) in terms of the coordinates \((y,\theta,\bar\theta)\) using that \(W_\alpha\) is gauge invariant.

A general abelian vector superfield \(V\) takes the form \[\begin{split} V(x,\theta,\bar\theta) &= C(x) + \theta\chi(x) + \bar\theta\bar\chi(x) +\theta\theta M(x) +\bar\theta\bar\theta\bar M(x) + (\theta\sigma^\mu\bar\theta) A_\mu(x) \\ & \quad + i(\theta\theta) \bar\theta\big(\bar\lambda(x) + \frac{1}{2}\bar\sigma^\mu\partial_\mu\chi(x)\big) - i(\bar\theta\bar\theta) \theta\big(\lambda(x) - \frac{1}{2}\sigma^\mu\partial_\mu\bar\chi(x)\big) \\ & \quad +\frac{1}{2}(\theta\theta)(\bar\theta\bar\theta)\big(D(x) - \frac{1}{2}\partial_\mu\partial^\mu C(x)\big) ~, \end{split}\] and we take the chiral superfield \(\Lambda\) and its conjugate \(\bar\Lambda\) to be given by \[\Lambda(y,\theta) = \phi(y,\theta) + \sqrt{2}\theta\psi(y) - \theta\theta F(y) ~, \qquad \bar\Lambda(\bar y,\bar \theta) = \bar \phi(\bar y,\bar \theta) + \sqrt{2}\bar\theta\bar\psi(y) - \bar\theta\bar\theta \bar F(\bar y) ~,\] or, in terms of the coordinates \((x,\theta,\bar\theta)\), \[\begin{split} \Lambda(x,\theta,\bar\theta) & = \phi(x) + i(\theta\sigma^\mu\bar\theta)\partial_\mu\phi(x) - \frac{1}{4}(\theta\theta)(\bar\theta\bar\theta)\partial_\mu\partial^\mu \phi(x) \\ & \quad + \sqrt{2}\theta\psi(x) - \frac{i}{\sqrt{2}}(\theta\theta)(\partial_\mu\psi(x)\sigma^\mu\bar\theta)-\theta\theta F(x) ~, \\ \bar \Lambda(x,\theta,\bar\theta) & = \bar \phi(x) - i(\theta\sigma^\mu\bar\theta)\partial_\mu\bar\phi(x) - \frac{1}{4}(\theta\theta)(\bar\theta\bar\theta)\partial_\mu\partial^\mu \bar\phi(x) \\ & \quad + \sqrt{2}\bar\theta\bar\psi(x) + \frac{i}{\sqrt{2}}(\bar\theta\bar\theta)(\theta\sigma^\mu\partial_\mu\bar\psi(x))-\bar\theta\bar\theta \bar F(x) ~. \end{split}\] Therefore, we have that \[\begin{split} V(x,\theta,\bar\theta) + \Lambda(x,\theta,\bar\theta) + \bar\Lambda(x,\theta,\bar\theta) & = C(x) + \phi(x) + \bar\phi(x) +\theta\big(\chi(x) + \sqrt{2}\psi(x)\big) +\bar\theta\big(\bar\chi(x) + \sqrt{2}\bar\psi(x)\big) \\ & \quad + \theta\theta \big(M(x) - F(x)\big) + \bar\theta\bar\theta\big(\bar M(x)- \bar F(x)\big) \\ & \quad + (\theta\sigma^\mu\bar\theta) \big(A_\mu(x) + i\partial_\mu\phi(x) - i\partial_\mu\bar\phi(x)\big) \\ & \quad + i(\theta\theta) \bar\theta\big(\bar\lambda(x) + \frac{1}{2}\bar\sigma^\mu\partial_\mu\chi(x) + \frac{1}{\sqrt{2}}\bar\sigma^\mu\partial_\mu\psi(x)\big) \\ & \quad - i(\bar\theta\bar\theta) \theta\big(\lambda(x) - \frac{1}{2}\sigma^\mu\partial_\mu\bar\chi(x) - \frac{1}{\sqrt{2}}\sigma^\mu\partial_\mu\bar\psi(x)\big) \\ & \quad +\frac{1}{2}(\theta\theta)(\bar\theta\bar\theta)\big(D(x) - \frac{1}{2}\partial_\mu\partial^\mu C(x) \\ & \hspace{100pt} - \frac{1}{2}\partial_\mu\partial^\mu \phi(x) - \frac{1}{2}\partial_\mu\partial^\mu \bar\phi(x)\big) ~, \end{split}\] hence the gauge transformations of the component fields are \[\begin{aligned} C & \to C + 2\mathop{\mathrm{Re}}\phi ~, & \qquad \chi & \to \chi + \sqrt{2}\psi ~, & \qquad M & \to M - F ~, \\ A_\mu & \to A_\mu - 2\partial_\mu \mathop{\mathrm{Im}}\phi ~, & \qquad \lambda & \to \lambda ~, & \qquad D & \to D ~. \end{aligned}\]

From the results of exercise 4.3, we have that the supersymmetry transformations of the component fields of an abelian vector superfield are \[\begin{aligned} \delta_{\varepsilon,\bar\varepsilon} C & = \varepsilon \chi + \bar\varepsilon\bar\chi ~, \qquad \delta_{\varepsilon,\bar\varepsilon} \chi = i(\sigma^\mu\bar\varepsilon)\big(\partial_\mu C -i A_\mu\big) + 2\varepsilon M ~, \qquad \delta_{\varepsilon,\bar\varepsilon} M = i\bar\varepsilon\big(\bar\lambda + \bar\sigma^\mu\partial_\mu\chi\big) ~, \\ \delta_{\varepsilon,\bar\varepsilon} A_\mu & = i\varepsilon\big(\sigma_\mu\bar\lambda +\frac{1}{2}(\sigma_\mu\bar\sigma^\nu+\sigma^\nu\bar\sigma_\mu)\partial_\nu\chi\big) +i\bar\varepsilon\big(\bar\sigma_\mu\lambda - \frac{1}{2}(\bar\sigma_\mu\sigma^\nu+\bar\sigma^\nu\sigma_\mu)\partial_\nu\bar\chi\big) \\ & = i\varepsilon\big(\sigma_\mu\bar\lambda +\partial_\mu\chi\big) +i\bar\varepsilon\big(\bar\sigma_\mu\lambda - \partial_\mu\bar\chi\big) ~, \\ \delta_{\varepsilon,\bar\varepsilon} \lambda & = \frac{1}{2}\sigma^\mu\partial_\mu\delta_{\varepsilon,\bar\varepsilon} \bar\chi - (\sigma^\mu\bar\varepsilon)\partial_\mu \bar M + \frac{1}{2}(\sigma^\nu\bar\sigma^\mu\varepsilon)\partial_\mu A_\nu +i \varepsilon \big(D-\frac{1}{2}\partial_\mu\partial^\mu C\big) \\ & = \frac{1}{2}\sigma^\mu\partial_\mu(i\bar\sigma^\nu\varepsilon(\partial_\nu C + i A_\nu) + 2\bar\varepsilon \bar M) - (\sigma^\mu\bar\varepsilon)\partial_\mu \bar M + \frac{1}{2}(\sigma^\nu\bar\sigma^\mu\varepsilon)\partial_\mu A_\nu +i \varepsilon \big(D-\frac{1}{2}\partial_\mu\partial^\mu C\big) \\ & = \frac{i}{4}(\sigma^\mu\bar\sigma^\nu+\sigma^\nu\bar\sigma^\mu)\varepsilon\partial_\mu\partial_\nu C - \frac{i}{2}\varepsilon \partial_\mu\partial^\mu C - \frac{1}{2}(\sigma^\mu\bar\sigma^\nu\varepsilon)(\partial_\mu A_\nu - \partial_\nu A_\mu) + i\varepsilon D \\ & = - \frac{1}{2}(\sigma^\mu\bar\sigma^\nu\varepsilon)(\partial_\mu A_\nu - \partial_\nu A_\mu) + i\varepsilon D ~, \\ \delta_{\varepsilon,\bar\varepsilon} D & = \frac{1}{2}\partial_\mu\partial^\mu\delta_{\varepsilon,\bar\varepsilon} C + i\varepsilon\sigma^\mu\partial_\mu\big(i\bar\lambda + \frac{i}{2}\bar\sigma^\nu\partial_\nu\chi\big) + i\bar\varepsilon\bar\sigma^\mu\partial_\mu\big(-i \lambda+\frac{i}{2}\sigma^\nu\partial_\nu\bar\chi\big) \\ & = - \varepsilon\sigma^\mu\partial_\mu\bar\lambda + \bar\varepsilon\bar\sigma^\mu\partial_\mu\lambda + \frac{1}{2}\varepsilon\partial_\mu\partial^\mu\chi + \frac{1}{2}\bar\varepsilon\partial_\mu\partial^\mu\bar\chi \\ & \quad - \frac{1}{4}\varepsilon(\sigma^\mu\bar\sigma^\nu+\sigma^\nu\bar\sigma^\mu)\partial_\mu\partial_\nu\chi - \frac{1}{4}\bar\varepsilon(\bar\sigma^\mu\sigma^\nu+\bar\sigma^\nu\sigma^\mu)\partial_\mu\partial_\nu\bar\chi \\ & = - \varepsilon\sigma^\mu\partial_\mu\bar\lambda + \bar\varepsilon\bar\sigma^\mu\partial_\mu\lambda ~, \end{aligned}\] where we have used that \(\sigma^\mu\bar\sigma^\nu+\sigma^\nu\bar\sigma^\mu = \bar\sigma^\mu\sigma^\nu+\bar\sigma^\nu\sigma^\mu = 2\eta^{\mu\nu}\mathbf{1}\). In the WZ gauge, \(C = \chi = M = 0\), the supersymmetry transformations simplify to \[\begin{aligned} \delta_{\varepsilon,\bar\varepsilon} C & = 0 ~, \qquad \delta_{\varepsilon,\bar\varepsilon} \chi = (\sigma^\mu\bar\varepsilon) A_\mu ~, \qquad \delta_{\varepsilon,\bar\varepsilon} M = i\bar\varepsilon\bar\lambda ~, \\ \delta_{\varepsilon,\bar\varepsilon} A_\mu & = i(\varepsilon\sigma_\mu\bar\lambda) +i(\bar\varepsilon\bar\sigma_\mu\lambda) ~, \\ \delta_{\varepsilon,\bar\varepsilon} \lambda & = - \frac{1}{2}(\sigma^\mu\bar\sigma^\nu\varepsilon)(\partial_\mu A_\nu - \partial_\nu A_\mu) + i\varepsilon D ~, \\ \delta_{\varepsilon,\bar\varepsilon} D & = - \varepsilon\sigma^\mu\partial_\mu\bar\lambda + \bar\varepsilon\bar\sigma^\mu\partial_\mu\lambda ~, \end{aligned}\] and we see explicitly that the WZ gauge is not supersymmetric. Recalling that the gauge transformations of the component fields are \[\begin{aligned} \delta_{\Lambda,\bar\Lambda} C & = 2\mathop{\mathrm{Re}}\phi ~, & \qquad \delta_{\Lambda,\bar\Lambda} \chi & = \sqrt{2}\psi ~, & \qquad \delta_{\Lambda,\bar\Lambda} M & = -F ~, \\ \delta_{\Lambda,\bar\Lambda} A_\mu & = -2\partial_\mu\mathop{\mathrm{Im}}\phi ~, & \qquad \delta_{\Lambda,\bar\Lambda} \lambda & = 0 ~, & \qquad \delta_{\Lambda,\bar\Lambda} D & = 0~, \end{aligned}\] it follows that to restore the WZ gauge we should set \[\phi = 0 ~, \qquad \psi = - \frac{1}{\sqrt{2}}(\sigma^\mu\bar\varepsilon)A_\mu ~, \qquad F = i \bar\varepsilon \bar\lambda ~.\] Therefore, the compensating gauge transformation needed to restore the WZ gauge is \[\begin{split} V \to V + \Lambda + \bar\Lambda ~, \qquad \Lambda(y,\theta) & = - (\theta\sigma^\mu\bar\varepsilon)A_\mu(y) - i(\theta\theta)(\bar\varepsilon \bar\lambda(y)) ~, \\ \Lambda(x,\theta,\bar\theta) & = - (\theta\sigma^\mu\bar\varepsilon)A_\mu(x) - i(\theta\sigma^\mu\bar\varepsilon)(\theta\sigma^\nu\bar\theta)\partial_\nu A_\mu(x) - i(\theta\theta)(\bar\varepsilon \bar\lambda(x)) \\ & = - (\theta\sigma^\mu\bar\varepsilon)A_\mu(x) - \frac{i}{2}(\theta\theta)(\bar\theta\bar\sigma^\mu\sigma^\nu \bar\varepsilon)\partial_\mu A_\nu(x) - i(\theta\theta)(\bar\varepsilon \bar\lambda(x)) ~, \end{split}\] such that, in the WZ gauge, we have \[\begin{aligned} (\delta_{\varepsilon,\bar\varepsilon}+\delta_{\Lambda,\bar\Lambda}) C & = 0 ~, \qquad (\delta_{\varepsilon,\bar\varepsilon}+\delta_{\Lambda,\bar\Lambda}) \chi = 0 ~, \qquad (\delta_{\varepsilon,\bar\varepsilon}+\delta_{\Lambda,\bar\Lambda}) M = 0 ~, \\ (\delta_{\varepsilon,\bar\varepsilon}+\delta_{\Lambda,\bar\Lambda}) A_\mu & = i(\varepsilon\sigma_\mu\bar\lambda) +i(\bar\varepsilon\bar\sigma_\mu\lambda) ~, \\ (\delta_{\varepsilon,\bar\varepsilon}+\delta_{\Lambda,\bar\Lambda}) \lambda & = - \frac{1}{2}(\sigma^\mu\bar\sigma^\nu\varepsilon)(\partial_\mu A_\nu - \partial_\nu A_\mu) + i\varepsilon D ~, \\ (\delta_{\varepsilon,\bar\varepsilon}+\delta_{\Lambda,\bar\Lambda}) D & = - \varepsilon\sigma^\mu\partial_\mu\bar\lambda + \bar\varepsilon\bar\sigma^\mu\partial_\mu\lambda ~. \end{aligned}\]

Since \(W_\alpha\) is gauge invariant, its general form is equal to its form in the WZ gauge, i.e. \(W_\alpha = - \frac{1}{4} \bar D^2 D_\alpha V = - \frac{1}{4} \bar D^2 D_\alpha V_{\text{WZ}}\). In terms of the coordinates \((y,\theta,\bar\theta)\) we have \(D_\alpha = \hat\partial_\alpha + 2i(\sigma^\mu\bar\theta)_\alpha \hat\partial_\mu\), \(\bar D_{\dot\alpha} = \hat{\bar\partial}_{\dot\alpha}\) and \[\begin{split} V_{\text{WZ}}(y,\theta,\bar\theta) & = (\theta\sigma^\mu\bar\theta) \big(A_\mu(y)-i (\theta\sigma^\nu\bar\theta)\hat\partial_\nu A_\mu\big) + i (\theta\theta) (\bar\theta\bar\lambda(y)) - i (\bar\theta\bar\theta)(\theta\lambda(y)) + \frac{1}{2} (\theta\theta) (\bar\theta\bar\theta) D(y) \\ & = (\theta\sigma^\mu\bar\theta) A_\mu(y) + i (\theta\theta) (\bar\theta\bar\lambda(y)) - i (\bar\theta\bar\theta)(\theta\lambda(y)) + \frac{1}{2} (\theta\theta) (\bar\theta\bar\theta) \big(D(y)-i \hat\partial_\mu A^\mu(y)\big) ~. \end{split}\] Therefore, \[\begin{split} D_\alpha V_{\text{WZ}}(y,\theta,\bar\theta) & = (\sigma^\mu\bar\theta)_\alpha A_\mu(y) + 2i (\sigma^\nu\bar\theta)_\alpha(\theta\sigma^\mu\bar\theta) \hat\partial_\nu A_\mu(y) + 2i \theta_\alpha (\bar\theta\bar\lambda(y)) - 2 (\sigma^\mu\bar\theta)_\alpha (\theta\theta) (\bar\theta\hat\partial_\mu \bar\lambda(y)) \\ & \quad - i (\bar\theta\bar\theta)\lambda_\alpha(y) + \theta_\alpha (\bar\theta\bar\theta) \big(D(y)-i \hat\partial_\mu A^\mu(y)\big) \\ & = (\sigma^\mu\bar\theta)_\alpha A_\mu(y) + i (\sigma^\nu\bar\sigma^\mu\theta)_{\alpha} (\bar\theta\bar\theta) \hat\partial_\nu A_\mu(y) -i \theta_\alpha (\bar\theta\bar\theta) \hat\partial_\mu A^\mu(y) \\ & \quad + 2i \theta_\alpha (\bar\theta\bar\lambda(y)) + (\theta\theta)(\bar\theta\bar\theta) (\sigma^\mu\hat\partial_\mu\bar\lambda(y))_{\alpha} - i (\bar\theta\bar\theta)\lambda_\alpha(y) + \theta_\alpha (\bar\theta\bar\theta) D(y) \\ & = (\sigma^\mu\bar\theta)_\alpha A_\mu(y) + \frac{i}{2} ((2\eta^{\mu\nu}\mathbf{1}- 4\sigma^{\mu\nu})\theta)_{\alpha} (\bar\theta\bar\theta) \hat\partial_\nu A_\mu(y) -i \theta_\alpha (\bar\theta\bar\theta) \hat\partial_\mu A^\mu(y) \\ & \quad + 2i \theta_\alpha (\bar\theta\bar\lambda(y)) + (\theta\theta)(\bar\theta\bar\theta) (\sigma^\mu\hat\partial_\mu\bar\lambda(y))_{\alpha} - i (\bar\theta\bar\theta)\lambda_\alpha(y) + \theta_\alpha (\bar\theta\bar\theta) D(y) \\ & = (\sigma^\mu\bar\theta)_\alpha A_\mu(y) + i (\sigma^{\mu\nu}\theta)_{\alpha} (\bar\theta\bar\theta) F_{\mu\nu}(y) \\ & \quad + 2i \theta_\alpha (\bar\theta\bar\lambda(y)) + (\theta\theta)(\bar\theta\bar\theta) (\sigma^\mu\hat\partial_\mu\bar\lambda(y))_{\alpha} - i (\bar\theta\bar\theta)\lambda_\alpha(y) + \theta_\alpha (\bar\theta\bar\theta) D(y) ~, \end{split}\] where \(F_{\mu\nu}(y) = \hat\partial_\mu A_\nu(y) - \hat\partial_\nu A_\mu(y)\) is the abelian field strength and we recall that \[\begin{aligned} \sigma^\mu\bar\sigma^\nu - \sigma^\nu\bar\sigma^\mu & = 4\sigma^{\mu\nu} ~, \qquad & \bar\sigma^\mu\sigma^\nu - \bar\sigma^\nu\sigma^\mu & = 4\bar\sigma^{\mu\nu} ~, \\ \sigma^\mu\bar\sigma^\nu + \sigma^\nu\bar\sigma^\mu & = 2\eta^{\mu\nu}\mathbf{1}~, \qquad & \bar\sigma^\mu\sigma^\nu + \bar\sigma^\nu\sigma^\mu & = 2\eta^{\mu\nu}\mathbf{1}~. \end{aligned}\] Now using that \(\hat{\bar\partial}^2(\bar\theta\bar\theta) = -4\) we find \[\begin{split} W_\alpha(y,\theta,\bar\theta) & = -\frac{1}{4} \bar D^2 D_\alpha V_{\text{WZ}}(y,\theta,\bar\theta) \\ & = - i \lambda_\alpha(y) + i (\sigma^{\mu\nu}\theta)_{\alpha} F_{\mu\nu}(y) + \theta_\alpha D(y) + (\theta\theta) (\sigma^\mu\hat\partial_\mu\bar\lambda(y))_{\alpha} ~. \end{split}\]

We are now in a position to construct supersymmetric actions for the abelian vector superfield. To match with standard conventions we also redefine \[\Lambda = - \frac{i}{2} \Delta ~, \qquad \Delta(y,\theta) = \phi(y) + \sqrt{2}\theta\psi(y) - \theta\theta F(y) ~,\] so that under supersymmetric gauge transformations \[V \to V - \frac{i}{2}\Delta + \frac{i}{2}\bar\Delta ~,\] and \[\begin{aligned} C & \to C - i \mathop{\mathrm{Re}}\phi ~, & \qquad A_\mu & \to A_\mu + i \partial_\mu \mathop{\mathrm{Im}}\phi ~, \\ M & \to M + \frac{i}{2} F ~, & \qquad D & \to D ~, \\ \chi & \to \chi - \frac{i}{\sqrt{2}} \psi ~, & \qquad \lambda & \to \lambda ~. \end{aligned}\]

The first type of term we consider is the kinetic term, which is constructed from the following superspace integral \[\int d^4 x d^2\theta \, W^\alpha W_\alpha ~.\] Computing the \(\mathcal{O}(\theta\theta)\) term in \(W^\alpha W_\alpha\) we find \[\begin{split} W^\alpha W_\alpha\big|_{\theta\theta} & = -2i \lambda \sigma^\mu\partial_\mu \bar\lambda + D^2 -\frac{1}{4}(\eta^{\mu\rho}\eta^{\nu\sigma} - \eta^{\nu\rho}\eta^{\mu\sigma} - i \epsilon^{\mu\nu\rho\sigma}) F_{\mu\nu} F_{\rho\sigma} \\ & = -2i \lambda \sigma^\mu\partial_\mu \bar\lambda + D^2-\frac{1}{2} F^{\mu\nu} F_{\mu\nu} + \frac{i}{4}\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu} F_{\rho\sigma} ~, \end{split}\] where we have used that \((\theta\sigma^{\mu\nu}\theta) = \frac{1}{4}(\theta\sigma^\mu\bar\sigma^\nu\theta - \theta\sigma^\nu\bar\sigma^\mu\theta) = 0\) and the identity \[(\sigma^{\mu\nu})^{\alpha\beta}(\sigma^{\rho\sigma})_{\beta\gamma} = \frac{1}{4}(\eta^{\mu\rho}\eta^{\nu\sigma} - \eta^{\nu\rho}\eta^{\mu\sigma} - i \epsilon^{\mu\nu\rho\sigma})\delta^\alpha_\gamma ~.\] Note that the \(\mathcal{O}(\theta\theta)\) term has the same form whether we evaluate the component fields at \(y^\mu\) or \(x^\mu\). The first three terms in \(W^\alpha W_\alpha\big|_{\theta\theta}\) are real, while the final term is imaginary. Introducing the complexified gauge coupling \[\tau = \frac{\theta}{2\pi} + \frac{4\pi i}{g^2} ~,\] where \(g\) is the gauge coupling and \(\theta\) is the theta angle, the kinetic term is given by \[\begin{split} \mathcal{S}_{\text{Maxwell}} & = \mathop{\mathrm{Im}}\Big(\int d^4 x d^2\theta \, \frac{\tau}{8\pi} W^\alpha W_\alpha\Big) \\ & = \int d^4x\, \big(\frac{1}{g^2}(-\frac{1}{4} F^{\mu\nu}F_{\mu\nu} - i\bar \lambda\bar\sigma^\mu\partial_\mu\lambda + \frac{1}{2} D^2) + \frac{\theta}{32\pi^2} \tilde F^{\mu\nu}F_{\mu\nu} \big) ~, \end{split}\] where \(\tilde F^{\mu\nu} = \frac{1}{2}\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}\) is the dual field strength. Here the kinetic term \(\mathcal{S}_{\text{Maxwell}}\) is written as an integral over chiral superspace, but the integrand is actually chirally exact \[W^\alpha W_\alpha = -\frac{1}{4} \bar D^2 (W^\alpha D_\alpha V) ~,\] hence strictly speaking it is a D-term.

The second type of term we consider is the Fayet-Iliopoulos (FI) term \[\mathcal{S}_{\text{FI}} = -2\xi \int d^4 x d^2\theta d^2\bar\theta \, V ~,\] where \(\xi\) is the FI parameter. This term is gauge invariant under \(V \to V -\frac{i}{2} \Delta + \frac{i}{2} \bar \Delta\) since \(\Delta\) and \(\bar \Delta\) are chiral and antichiral superfields respectively. The existence of this term is a peculiarity of abelian supersymmetric gauge theories. In terms of component fields we have \[\mathcal{S}_{\text{FI}} = -\xi \int d^4x \, D ~.\]

The third type of term we consider describes the coupling to matter fields. Let us consider a chiral superfield \(\Phi\) that transforms under the \(\mathrm{U}(1)\) gauge symmetry with charge \(q\), \([\Phi]_{_Q} = q\). Under supersymmetric gauge transformations we have \[\Phi \to e^{i q\Delta} \Phi ~, \qquad \bar\Phi \to \bar\Phi e^{-i q\bar \Delta} ~, \qquad V \to V - \frac{i}{2}\Delta +\frac{i}{2}\bar\Delta ~.\] It is now straightforward to construct a gauge invariant kinetic term for \(\Phi\) \[\mathcal{S}_{\text{matter}} = \int d^4x d^2\theta d^2\bar\theta \, \bar\Phi e^{2qV} \Phi ~.\] Since this term is gauge invariant, we can compute it in the WZ gauge. We have \[\begin{split} e^{2 q V_{\text{WZ}}} & = 1 + 2 q V_{\text{WZ}} + 2q^2 V_{\text{WZ}}^2 \\ & = 1+ 2q(\theta\sigma^\mu\bar\theta) A_\mu + 2iq(\theta\theta)(\bar\theta\bar\lambda) - 2iq (\bar\theta\bar\theta)(\theta\lambda) + (\theta\theta)(\bar\theta\bar\theta)(qD+q^2A_\mu A^\mu) ~, \end{split}\] hence, up to total derivatives, \[\begin{split} \bar\Phi e^{2qV}\Phi\big|_{\theta\theta\bar\theta\bar\theta} & = \bar D_\mu \bar\phi D^\mu \phi - i \bar \psi \bar\sigma^\mu D_\mu \psi + \bar F F + i q\sqrt{2} \bar\phi\lambda\psi - i q\sqrt{2} \bar\psi\bar\lambda\phi + q \bar\phi D \phi ~, \end{split}\] where the gauge covariant derivative acting on a field of charge \(q\) is given by \[D_\mu = \partial_\mu - i q A_\mu ~.\] Therefore, \[\mathcal{S}_{\text{matter}} = \int d^4x\,\big(\bar D_\mu \bar\Phi D^\mu \Phi - i \bar \psi \bar\sigma^\mu D_\mu \psi + \bar F F + i q\sqrt{2} \bar\phi\lambda\psi - i q\sqrt{2} \bar\psi\bar\lambda\phi + q \bar\phi D \phi\big) ~,\] up to total derivatives. We may also have superpotential terms \[\begin{split} \mathcal{S}_{\text{W}} & = \int d^4x d^2\theta \, W(\Phi) + \int d^4x d^2\bar\theta \, \bar W(\bar\Phi) ~, \end{split}\] where \(W(\Phi)\) is a gauge invariant holomorphic function of \(\Phi\), i.e. \([W]_{_Q} = 0\).

From these four types of terms we can construct abelian supersymmetric gauge theories. We take \(r\) abelian vector superfields \(V_a\), \(a=1,\dots,r\), and \(N\) chiral superfields \(\Phi^i\), \(i=1,\dots,N\), with charges \([\Phi^i]_{_{Q_a}} = q^i_a\). We can then write down the following (renormalisable) gauge invariant action \[\begin{aligned} \mathcal{S}& = \mathcal{S}_{\text{Maxwell}} + \mathcal{S}_{\text{FI}} + \mathcal{S}_{\text{matter}} + \mathcal{S}_{\text{W}} ~, \\ \mathcal{S}_{\text{Maxwell}} & = \sum_{a=1}^r \mathop{\mathrm{Im}}\Big(\int d^4x d^2\theta \, \frac{\tau_a}{8\pi} W_a^{\alpha} W_{a\alpha} \Big) ~, \qquad \tau_a = \frac{\theta_a}{2\pi} + \frac{4\pi i}{g_a^2} ~, \\ \mathcal{S}_{\text{FI}} & = -2 \sum_{a=1}^r \xi_a \int d^4x d^2\theta d^2\bar\theta \, V_a ~, \\ \mathcal{S}_{\text{matter}} & = \sum_{i=1}^N \int d^4x d^2\theta d^2\bar\theta \, \bar \Phi_i e^{2\sum_{a=1}^r q_a^i V_a} \Phi^i ~, \\ \mathcal{S}_{\text{W}} & = \int d^4x d^2\theta \, W(\Phi) + \int d^4x d^2\bar \theta \, \bar W(\bar\Phi) ~, \end{aligned}\] where the superpotential \(W(\Phi)\) is a gauge invariant holomorphic function of \(\Phi^i\), i.e. \([W]_{_{Q_a}} = 0\) for all \(a=1,\dots r\).

In order to investigate the physics of these theories, let us construct the scalar potential, which comes from integrating out the auxiliary fields \(D_a\), \(F^i\) and \(\bar F_i\). The terms involving these fields are \[\mathcal{S}_{\text{aux}} = \int d^4x \,\big( \sum_{a=1}^r (\frac{1}{2g_a^2} (D_a)^2 - \xi_a D_a + D_a \sum_{i=1}^Nq_a^i \bar\phi_i \phi^i) + \sum_{i=1}^N(\bar F_i F^i - \partial_i W(\phi) F^i - \bar\partial^i \bar W(\bar\phi) \bar F_i) \big) ~.\] Therefore, integrating out the auxiliary fields gives \[\begin{split} \bar F_i & = \partial_i W(\phi) ~, \qquad F^i = \bar\partial^i \bar W(\bar\phi) ~, \\ D_a & = -g_a^2 \big(\sum_{i=1}^N q_a^i\bar\phi_i \phi^i - \xi_a\big) = -g_a^2 (\mu_a(\phi,\bar\phi) - \xi_a) ~, \end{split}\] where we have defined the so-called moment map \[\mu_a(\phi,\bar\phi) = \sum_{i=1}^N q_a^i \bar\phi_i\phi^i ~.\] Substituting back into \(\mathcal{S}_{\text{aux}}\) we find the scalar potential \[\begin{split} V(\phi,\bar\phi) & = \sum_{i=1}^N \bar F_i F^i + \sum_{a=1}^r \frac{1}{2g_a^2} (D_a)^2 \\ & = \sum_{i=1}^N \partial_i W(\phi) \bar\partial^i \bar W(\bar\phi) + \sum_{a=1}^r \frac{g_a^2}{2}(\mu_a(\phi,\bar\phi)-\xi_a)^2 ~, \end{split}\] which we see has the general structure of |F-term|\(^2\) + (D-term)\(^2\).

Recalling that only the scalar fields can have non-vanishing vacuum expectation values, now that we have the scalar potential we can compute the moduli space of supersymmetric vacua. For supersymmetric vacua the scalar potential should vanish. Therefore, since \(V(\phi,\bar\phi)\) is a sum of non-negative terms, each term should vanish separately, hence \[\mathcal{M} = \Big\{(\phi,\bar\phi) ~:~ \bar F_i = \partial_i W(\phi) = 0 ~~\text{and}~~ D_a = -g_a^2 (\mu_a(\phi,\bar\phi) - \xi_a) = 0 \Big\} \Big/ \mathrm{U}(1)^r ~,\] where all fields are understood to be given by their vacuum expectation values. We have quotiented out by the action of the gauge group \(\mathrm{U}(1)^r\) \[\phi^i \sim e^{i\sum_{a=1}^r q_a^i \alpha_a} \phi_i ~,\] since configurations related by gauge transformations are physically equivalent.

For vanishing FI parameters, it can then be shown that \(\mathcal{M}\) has the complex algebraic description \[\begin{equation} \label{eq:vevcad} \mathcal{M} = \Big\{ \phi^i \in \mathbb{C}^N ~:~ \partial_i W(\phi) = 0 \Big\} \Big/ (\mathbb{C}^*)^r ~, \end{equation}\] where \((\mathbb{C}^*)^r\) is the complexified gauge group. \(\mathbb{C}^* = \mathbb{C}\backslash\{0\}\) denotes the complex plane with the origin removed. That is, setting \(D_a = 0\) and quotienting out by the \(\mathrm{U}(1)^r\) gauge group is equivalent to quotienting out by the complexified gauge group \((\mathbb{C}^*)^r\) \[\phi^i \sim \big(\prod_{a=1}^r \lambda_a^{q_a^i} \big) \phi^i ~, \qquad \lambda_a \in \mathbb{C}^* ~,\] where \(\lambda_a\) can be thought of as the \(\mathcal{O}(1)\) term of the chiral superfield \(e^{i\Delta_a}\). While we are no longer enforcing \(r\) real constraints (\(D_a = 0\)), this is compensated by introducing \(r\) additional gauge degrees of freedom. In order to prove that \(\mathcal{M}\) can be written in this way, we need to show that any configuration of \(\phi^i\) satisfying \(\partial_i W(\phi) = 0\) can be bought to a configuration that also satisfies \(D_a = 0\). This is except for the configuration \(\phi^i = 0\), which is gauge invariant and already satisfies \(D_a = 0\). Moreover, this should be possible to do in a unique way up to the original \(\mathrm{U}(1)^r\) gauge symmetry. An outline of the proof goes as follows. We first note that, since \(W(\phi)\) is a gauge invariant holomorphic function of \(\phi^i\), the condition \(\partial_i W(\phi)\) is also invariant under the complexified gauge group \((\mathbb{C}^*)^r\). Under complexified gauge transformations we have \[D_a = -g_a^2 \sum_{i=1}^N q_a^i \bar\phi_i\phi^i \to -g_a^2\sum_{i=1}^N q_a^i \big(\prod_{b=1}^r (\bar\lambda_b\lambda_b)^{q_b^i}\big) \bar\phi_i \phi^i ~.\] Seeing as \(D_a\) only depends on \(\phi^i\) through \(\bar\phi_i\phi^i\) it is invariant under the original \(\mathrm{U}(1)^r\) gauge symmetry. Therefore, setting \(D_a = 0\) we find \(r\) real equations for the \(r\) real and positive variables \(\bar\lambda_a\lambda_a\). It is then possible to show that under natural physical assumptions this system of equations indeed has a unique solution. The complex algebraic description is useful because it implies that the moduli space of supersymmetric vacua is parametrised by the vacuum expectation values of gauge invariant chiral operators, subject to algebraic conditions. These operators have the structure of a ring, called the chiral ring.

Let us conclude with an example of an abelian supersymmetric gauge theory, supersymmetric quantum electrodynamics (SQED) with \(N_f\) flavours. We have a abelian vector superfield with \(\mathrm{U}(1)\) gauge group. In addition, we have \(2N_f\) matter fields: \(Q^i\), with charge \(1\) transforming in the antifundamental representation of the \(\mathrm{SU}(N_f)\) flavour symmetry, and \(\tilde Q_i\), with charge \(-1\) transforming in the fundamental representation of the \(\mathrm{SU}(N_f)\) flavour symmetry, where \(i=1,\dots N_f\). Setting the FI parameter to zero, we would now like to show that \(\mathcal{M}\) has the complex algebraic description \(\eqref{eq:vevcad}\). To do so we need to check that the equation \[D = - g^2 \sum_{i=1}^N \big( e^{2\chi} \bar \phi_{Qi} \phi^{Qi} - e^{-2\chi} \bar \phi^{\tilde Q i} \phi_{\tilde Q i}\big) = 0~,\] with \(\phi^{Qi} \neq 0\) and \(\phi^{\tilde Qj} \neq 0\) for some \(i\) and \(j\), has a unique solution for \(\chi \in \mathbb{R}\), where we have set \(\bar\lambda \lambda = e^{2\chi}\). If \(\phi^{Qi} = 0\) or \(\phi^{\tilde Qi} = 0\) for all \(i\), this unique solution is given by \(\chi \to \infty\) and \(\chi \to -\infty\) respectively. Otherwise, taking \(\chi \to \pm \infty\) we have \(D \to \mp \infty\). Moreover, \[\frac{\partial D}{\partial \chi} = -2 g^2 \sum_{i=1}^N \big( e^{2\chi} \bar \phi_{Qi} \phi^{Qi} + e^{-2\chi} \bar \phi^{\tilde Q i} \phi_{\tilde Q i}\big) ~,\] which is strictly negative. Therefore, \(D\) is a monotonically decreasing function of \(\chi\) satisfying \(D \to \mp \infty\) as \(\chi \to \pm \infty\), hence equals zero exactly once.

The field content of an abelian supersymmetric gauge theory consists of \(r\) abelian vector superfields \(V_a\), \(a=1,\dots,r\), and \(N\) chiral superfields \(\Phi^i\), \(i=1,\dots,N\), with charges \([\Phi^i]_{_{Q_a}} = q^i_a\) under the \(\mathrm{U}(1)^r\) gauge group.

Write down the component field expansion of the most general supersymmetric action \[\begin{split} \mathcal{S}& = \mathcal{S}_{\text{Maxwell}} + \mathcal{S}_{\text{FI}} + \mathcal{S}_{\text{matter}} + \mathcal{S}_{\text{W}} ~, \\ \mathcal{S}_{\text{Maxwell}} & = \sum_{a=1}^r \mathop{\mathrm{Im}}\Big(\int d^4x d^2\theta \, \frac{\tau_a}{8\pi} W_a^{\alpha} W_{a\alpha} \Big) ~, \qquad \tau_a = \frac{\theta_a}{2\pi} + \frac{4\pi i}{g_a^2} ~, \\ \mathcal{S}_{\text{FI}} & = -2 \sum_{a=1}^r \xi_a \int d^4x d^2\theta d^2\bar\theta \, V_a ~, \\ \mathcal{S}_{\text{matter}} & = \sum_{i=1}^N \int d^4x d^2\theta d^2\bar\theta \, \bar \Phi_i e^{2\sum_{a=1}^r q_a^i V_a} \Phi^i ~, \\ \mathcal{S}_{\text{W}} & = \int d^4x d^2\theta \, W(\Phi) + \int d^4x d^2\bar \theta \, \bar W(\bar\Phi) ~, \end{split}\] where the superpotential \(W(\Phi)\) is a gauge invariant holomorphic function of \(\Phi^i\), i.e. \([W]_{_{Q_a}} = 0\) for all \(a=1,\dots r\).

Integrate out the auxiliary fields \(F^i\), \(\bar F_i\) and \(D_a\) in the chiral and abelian vector superfields, and derive the scalar potential \[\begin{split} V(\phi,\bar\phi) & = \sum_{i=1}^N \bar F_i F^i + \sum_{a=1}^r \frac{1}{2g_a^2} (D_a)^2 \\ & = \sum_{i=1}^N \partial_i W(\phi) \bar\partial^i \bar W(\bar\phi) + \sum_{a=1}^r \frac{g_a^2}{2}(\mu_a(\phi,\bar\phi)-\xi_a)^2 ~, \qquad \mu_a(\phi,\bar\phi) = \sum_{i=1}^N q_a^i \bar\phi_i\phi^i ~. \end{split}\]

Up total derivatives, the component field expansion of the most general supersymmetric action for \(r\) vector superfields and \(N\) chiral superfields is \[\begin{split} \mathcal{S}& = \sum_{a=1}^r \int d^4x\, \big(\frac{1}{g_a^2}(-\frac{1}{4} F_a^{\mu\nu}F_{a\mu\nu} - i\bar \lambda_a\bar\sigma^\mu\partial_\mu\lambda_a + \frac{1}{2} (D_a)^2) + \frac{\theta_a}{32\pi^2} \tilde F_a^{\mu\nu}F_{a\mu\nu} - \xi_a D_a \big) \\ & \quad + \sum_{i=1}^N \int d^4x\,\big(\bar D_\mu \bar\phi_i D^\mu \phi^i - i \bar \psi_i \bar\sigma^\mu D_\mu \psi^i + \bar F_i F^i + \sum_{a=1}^r q_a^i (i \sqrt{2} \bar\phi_i\lambda_a\psi^i - i \sqrt{2} \bar\psi_i\bar\lambda_a\phi^i + \bar\phi_i D_a \phi^i)\big) \\ & \quad - \sum_{i=1}^N \int d^4 x \, \big(\partial_i W(\phi) F^i - \bar\partial^i \bar W(\bar \phi) \bar F_i\big) -\frac{1}{2} \sum_{i=1}^N \sum_{j=1}^N \int d^4 x \, \big(\partial_i\partial_j W(\phi) \psi^i\psi^j + \bar\partial^i\bar\partial^j \bar W(\bar \phi) \bar\psi_i\bar\psi_j\big) ~, \end{split}\] where \[D_\mu\phi^i = \partial_\mu\phi^i - i \sum_{a=1}^r q_a^i A_{a\mu} \phi^i ~, \\ \qquad D_\mu\psi^i = \partial_\mu\psi^i - i \sum_{a=1}^r q_a^i A_{a\mu} \psi^i ~.\] The terms involving the auxiliary fields are \[\mathcal{S}_{\text{aux}} = \int d^4x \,\big( \sum_{a=1}^r (\frac{1}{2g_a^2} (D_a)^2 - \xi_a D_a + D_a \sum_{i=1}^Nq_a^i \bar\phi_i \phi^i) + \sum_{i=1}^N(\bar F_i F^i - \partial_i W(\phi) F^i - \bar\partial^i \bar W(\bar\phi) \bar F_i) \big) ~.\] Varying with respect to \(F^i\), \(\bar F_i\) and \(D_a\) we find \[\bar F_i = \partial_i W(\phi) ~, \qquad F^i = \bar\partial^i \bar W(\bar\phi) ~, \qquad D_a = - g_a^2 \big(\sum_{i=1}^N q_a^i \bar\phi_i \phi^i - \xi_a\big) ~.\] Substituting back into \(\mathcal{S}_{\text{aux}}\) we find \[\mathcal{S}_{\text{aux}} = - \int d^4x \,\big( \sum_{a=1}^r \frac{1}{2g_a^2} (D_a)^2 + \sum_{i=1}^N \bar F_i F^i \big) ~,\] where the auxiliary fields are understood to solve their equations of motion, hence are functions of the scalar fields. These are all the terms in the action that involve only scalar fields without derivatives. Therefore, the scalar potential is indeed given by \[\begin{split} V(\phi,\bar\phi) & = \sum_{i=1}^N \bar F_i F^i + \sum_{a=1}^r \frac{1}{2g_a^2} (D_a)^2 \\ & = \sum_{i=1}^N \partial_i W(\phi) \bar\partial^i \bar W(\bar\phi) + \sum_{a=1}^r \frac{g_a^2}{2}(\mu_a(\phi,\bar\phi)-\xi_a)^2 ~, \qquad \mu_a(\phi,\bar\phi) = \sum_{i=1}^N q_a^i \bar\phi_i\phi^i ~. \end{split}\]

Supersymmetric quantum electrodynamics (SQED) with one flavour is a supersymmetric \(\mathrm{U}(1)\) gauge theory with matter fields, i.e. chiral superfields, \(Q\) of charge \(1\) and \(\tilde Q\) of charge \(-1\). Its action is given by \[\begin{split} \mathcal{S}& = \mathop{\mathrm{Im}}\Big(\int d^4 x d^2\theta \, \frac{\tau}{8\pi} W^\alpha W_\alpha \Big) + \int d^4x d^2\theta d^2\bar\theta \, \big(\bar Q e^{2V} Q + \bar{\tilde{Q}} e^{-2V} \tilde{Q} - 2\xi V\big) \\ & \quad + \int d^4 x d^2\theta \, m \tilde{Q} Q + \int d^4x d^2\bar\theta \, \bar m \bar{\tilde{Q}} \bar{Q} ~, \end{split}\] where \(m\) is a complex mass parameter, \(\tau = \frac{4\pi i}{g^2} + \frac{\theta}{2\pi}\) is the complexified gauge coupling, \(\xi\) is the real Fayet-Iliopoulos (FI) parameter, and \(W_\alpha = - \frac{1}{4} \bar D^2 D_\alpha V\) is the gaugino (or photino) superfield.

Derive the scalar potential of this theory in terms of the component fields of the chiral superfields \(Q\) and \(\tilde{Q}\). Determine the moduli space of supersymmetric vacua for

\(m = 0\), \(\xi = 0\);

\(m = 0\), \(\xi \neq 0\);

\(m \neq 0\), \(\xi = 0\);

\(m \neq 0\), \(\xi \neq 0\).

Determine the vacuum expectation values of the gauge invariant chiral operator \(M = \tilde{Q} Q\) in these supersymmetric vacua. You may use that unbroken supersymmetry implies that \(\langle XY \rangle = \langle X\rangle\langle Y \rangle\) for any two chiral superfields \(X\) and \(Y\).

The component field expansion of the chiral superfields \(Q\) and \(\tilde{Q}\) is \[Q = \phi^Q + \sqrt{2} \theta\psi^Q - \theta\theta F^Q ~, \qquad \tilde Q = \phi^{\tilde Q} + \sqrt{2} \theta\psi^{\tilde Q} - \theta\theta F^{\tilde Q} ~,\] The scalar potential is then given by \[V(\phi^Q, \phi^{\tilde{Q}}, \bar\phi_Q, \bar\phi_{\tilde{Q}}) = \bar F_Q F^Q + \bar F_{\tilde{Q}} F^{\tilde{Q}} + \frac{1}{2g^2} D^2 ~,\] where \[\begin{aligned} F^Q & = \partial_{\bar {\phi}_Q}(\bar m \bar\phi_{\tilde{Q}} \bar\phi_Q) = \bar m \bar\phi_{\tilde{Q}} ~, \qquad & \bar F_Q & = \partial_{\phi^Q}(m \phi^{\tilde Q}\phi^{Q}) = m \phi^{\tilde{Q}} ~, \\ F^{\tilde{Q}} & = \partial_{\bar{\phi}_{\tilde{Q}}}(\bar m \bar\phi_{\tilde{Q}} \bar\phi_Q) = \bar m \bar\phi_{Q} ~, \qquad & \bar F_{\tilde{Q}} & = \partial_{\phi^{\tilde{Q}}}(m \phi^{\tilde Q} \phi^{Q}) = m \phi^Q ~, \\ D & = -g^2 (\bar \phi_Q \phi^Q - \bar\phi_{\tilde{Q}} \phi^{\tilde Q} - \xi) ~. && \end{aligned}\] Therefore, \[V(\phi^Q, \phi^{\tilde{Q}}, \bar\phi_Q, \bar\phi_{\tilde{Q}}) = \bar m m(|\phi^Q|^2 + |\phi^{\tilde Q})|^2) + \frac{g^2}{2}(|\phi^Q|^2 - |\phi^{\tilde Q}|^2 - \xi)^2 ~.\]

To determine the moduli space of supersymmetric vacua we solve \(\bar F_Q = \bar F_{\tilde{Q}} = D = 0\) and quotient out by the \(\mathrm{U}(1)\) gauge symmetry \[\begin{aligned} Q & \to e^{i\alpha(x)} Q ~, \qquad & \bar Q & \to e^{-i\alpha(x)} \bar Q ~, \\ \tilde{Q} & \to e^{-i\alpha(x)} \tilde{Q} ~, \qquad & \bar{\tilde{Q}} & \to e^{i\alpha(x)} \bar{\tilde{Q}} ~. \end{aligned}\]

For \(m = 0\), \(\xi = 0\), we have \(\bar F_Q = \bar F_{\tilde{Q}} = 0\) and \(D = -g^2 (|\phi^Q|^2 - |\phi^{\tilde Q}|^2)\). Solving \(D = 0\), we find \(|\phi^Q|^2 = |\phi^{\tilde Q}|^2\). We can use the \(\mathrm{U}(1)\) gauge symmetry to fix \(\arg \phi^Q = \arg \phi^{\tilde{Q}}\), hence \(\phi^Q = \phi^{\tilde{Q}}\). Therefore, \[\mathcal{M} = \Big\{\phi^Q, \phi^{\tilde{Q}} ~:~ \phi^Q = \phi^{\tilde{Q}}\Big\} = \mathbb{C}~.\] The vacuum expectation value of \(M = \tilde{Q} Q\) is \(\langle M \rangle = \langle \tilde{Q}\rangle \langle Q\rangle = \langle \phi^{\tilde{Q}}\rangle\langle \phi^Q \rangle\), hence takes all complex values on \(\mathcal{M}\).

For \(m = 0\), \(\xi \neq 0\), we have \(\bar F_Q = \bar F_{\tilde{Q}} = 0\) and \(D = -g^2 (|\phi^Q|^2 - |\phi^{\tilde Q}|^2 - \xi)\). Solving \(D = 0\), we find \(|\phi^Q|^2 = |\phi^{\tilde Q}|^2 + \xi\). We can again use the \(\mathrm{U}(1)\) gauge symmetry to fix \(\arg \phi^Q = \arg \phi^{\tilde{Q}}\). Then for \(\xi > 0\), \(\phi^Q\) is determined in terms of \(\tilde\phi^Q\), while for \(\xi < 0\), \(\tilde\phi^Q\) is determined in terms of \(\phi^Q\). In particular, we have \[\begin{split} \xi > 0 ~:~ \qquad \mathcal{M} & = \Big\{\phi^Q, \phi^{\tilde{Q}} ~:~ \phi^Q = \sqrt{1+\frac{\xi}{|\phi^{\tilde Q}|^2}} \phi^{\tilde{Q}}\Big\} = \mathbb{C}~, \\ \xi < 0 ~:~ \qquad \mathcal{M} & = \Big\{\phi^Q, \phi^{\tilde{Q}} ~:~ \tilde{\phi}_Q = \sqrt{1-\frac{\xi}{|\phi^Q|^2}}\phi^Q\Big\} = \mathbb{C}~. \end{split}\] The vacuum expectation value of \(M = \tilde{Q} Q\) is \(\langle M \rangle = \langle \tilde{Q}\rangle \langle Q\rangle = \langle \phi^{\tilde{Q}}\rangle\langle \phi^Q \rangle\), hence takes all complex values on \(\mathcal{M}\).

For \(m \neq 0\), \(\xi = 0\), we have \(\bar F_Q = m \phi^{\tilde{Q}}\), \(\bar F_{\tilde{Q}} = m \phi^Q\) and \(D = -g^2 (|\phi^Q|^2 - |\phi^{\tilde Q}|^2)\). Solving \(\bar F_Q = \bar F_{\tilde{Q}} = 0\) gives \(\phi^Q = \phi^{\tilde{Q}} = 0\), for which \(D = 0\). The \(\mathrm{U}(1)\) gauge symmetry has a trivial action on \(\phi^Q = \phi^{\tilde{Q}} = 0\). Therefore, \[\mathcal{M} = \Big\{\phi^Q = 0, \phi^{\tilde{Q}} = 0 \Big\} = \cdot ~,\] i.e. a single point. The vacuum expectation value of \(M = \tilde{Q} Q\) is \(\langle M \rangle = \langle \tilde{Q}\rangle \langle Q\rangle = \langle \phi^{\tilde{Q}}\rangle\langle \phi^Q \rangle = 0\).

For \(m \neq 0\), \(\xi \neq 0\), we have \(\bar F_Q = m \phi^{\tilde{Q}}\), \(\bar F_{\tilde{Q}} = m \phi^Q\) and \(D = -g^2 (|\phi^Q|^2 - |\phi^{\tilde Q}|^2 -\xi)\). Solving \(\bar F_Q = \bar F_{\tilde{Q}} = 0\) gives \(\phi^Q = \phi^{\tilde{Q}} = 0\), for which \(D = g^2\xi \neq 0\). Therefore, there are no supersymmetric vacua \[\mathcal{M} = \emptyset ~,\] and supersymmetry is broken spontaneously.

Let us now turn to non-abelian supersymmetric gauge theories. In this discussion we will mainly focus on the differences compared to the abelian theories discussed in subsection 6.1. We start by considering a non-abelian supersymmetric gauge theory with compact gauge group \(\mathrm{G}\). We have a non-abelian vector superfield \(V = V_a T^a\), where \(T^a\), \(a=1,\dots,\dim\mathrm{G}\), are the generators in the adjoint representation of \(\mathrm{G}\), and a chiral superfield \(\Phi\), which transforms in the representation \(R\) of \(\mathrm{G}\) with generators \(T_{{\scriptscriptstyle{R}}}^a\). \(\bar \Phi_i\) transforms in the complex conjugate representation \(\bar R\), with generators \(T^a_{\bar{{\scriptscriptstyle{R}}}} = - (T^a_{\scriptscriptstyle{R}})^\mathrm{t}\). Under non-abelian supersymmetric gauge transformations \[e^{2V} \to e^{i\bar\Delta} e^{2V} e^{-i\Delta} ~, \qquad \Phi \to e^{i\Delta_{\scriptscriptstyle{R}}} \Phi ~, \qquad \bar\Phi \to \bar\Phi e^{-i\bar\Delta_{{\scriptscriptstyle{R}}}} ~, \qquad \Delta = \Delta_a T^a ~, \qquad \Delta_{{\scriptscriptstyle{R}}} = \Delta_a T_{\scriptscriptstyle{R}}^a ~.\] For a \(\mathrm{U}(1)\) gauge group under which \(\Phi\) has charge \(q\), we can take \(T_q = q \mathbf{1}\) as the generator in the charge \(q\) representation. The gauge invariant kinetic term for the chiral superfield \(\Phi\) is \[\mathcal{S}_{\text{matter}} = \int d^4x d^2\theta d^2\bar\theta \, \bar\Phi e^{2V_{\scriptscriptstyle{R}}} \Phi ~, \qquad V_{\scriptscriptstyle{R}} = V_a T_{{\scriptscriptstyle{R}}}^a ~,\] such that, under supersymmetric gauge transformations \[e^{2V_{\scriptscriptstyle{R}}} \to e^{i\bar\Delta_{\scriptscriptstyle{R}}} e^{2V_{\scriptscriptstyle{R}}} e^{-i\Delta_{\scriptscriptstyle{R}}} ~.\] More explicitly, we have \[\Phi^i \to (e^{i\Delta_a T_{{\scriptscriptstyle{R}}}^a})^i{}_j \Phi^j ~, \qquad \bar\Phi_i \to \bar\Phi_i (e^{-i\bar \Delta_a T_{{\scriptscriptstyle{R}}}^a})^i{}_j ~, \qquad \bar\Phi e^{2V_{\scriptscriptstyle{R}}} \Phi = \bar\Phi_i (e^{2V_{\scriptscriptstyle{R}}} )^i{}_j \Phi^j ~,\] where \(i = 1,\dots,\dim{R}\).

The non-abelian gaugino superfield (or gauge covariant super field strength) is \[W_\alpha = -\frac{1}{8} \bar D^2(e^{-2V} D_\alpha e^{2V}) ~, \qquad \bar W_{\dot\alpha} = \frac{1}{8} D^2(e^{2V} \bar D_{\dot\alpha} e^{-2V}) ~.\] Since \(e^{-2V} D_\alpha e^{2V} = 2 D_\alpha V + \dots\) and \(e^{2V} \bar D_{\dot\alpha} e^{-2V} = -2D_{\dot\alpha} V + \dots\) where the additional terms all involve commutators, we recover the abelian gaugino superfield for abelian gauge groups \(\mathrm{G}\). Under gauge transformations \[\begin{split} W_\alpha & \to -\frac{1}{8}\bar D^2(e^{i\Delta} e^{-2V} e^{-i\bar\Delta} D_\alpha (e^{i\bar\Delta} e^{2V} e^{-i \Delta})) = -\frac{1}{8} e^{i\Delta} \bar D^2 (e^{-2V} D_\alpha(e^{2V} e^{-i\Delta})) \\ & \qquad = -\frac{1}{8} e^{i\Delta} \bar D^2(e^{-2V}(D_\alpha e^{2V})e^{-i\Delta}) = -\frac{1}{8} e^{i\Delta} \bar D^2(e^{-2V}(D_\alpha e^{2V}))e^{-i\Delta} \\ & \qquad = e^{i\Delta} W_\alpha e^{-i\Delta} ~, \end{split}\] where we have used that \(\bar D_{\dot\alpha} e^{\pm i\Delta} = D_\alpha e^{\pm i\bar\Delta} = 0\) and \[\bar D^2 D_\alpha e^{-i\Delta} = -\bar D^{\dot\alpha} \bar D_{\dot\alpha} D_\alpha e^{-i\Delta} = -\bar D^{\dot\alpha}\{\bar D_{\dot\alpha}, D_\alpha\} e^{-i\Delta} = -2i (\sigma^\mu)_{\alpha\dot\alpha} \bar D^{\dot\alpha} \partial_\mu e^{-i\Delta} = 0 ~,\] since \(\Delta\) and \(\bar\Delta\) are chiral and antichiral superfields respectively. Therefore, we see that \(W_\alpha\) is gauge covariant. To compute \(W_\alpha\) in the WZ gauge we expand the exponentials and use that \(V_{\text{WZ}}^3 = (D_\alpha V_{\text{WZ}}) V_{\text{WZ}}^2 = V_{\text{WZ}} (D_\alpha V_{\text{WZ}}) V_{\text{WZ}} = V_{\text{WZ}}^2 (D_\alpha V_{\text{WZ}}) = 0\). Doing so we find \[W_{\text{WZ}}{}_\alpha = - \frac{1}{4} \bar D^2(D_\alpha V_{\text{WZ}} + [D_\alpha V_{\text{WZ}}, V_{\text{WZ}}] )~,\] the component field expansion of which is \[W_{\text{WZ}}{}_\alpha(y,\theta) = - i \lambda_\alpha(y) + \theta_\alpha D(y) + i (\sigma^{\mu\nu}\theta)_\alpha F_{\mu\nu}(y) + (\theta\theta)(\sigma^\mu D_\mu \bar \lambda(y))_\alpha ~,\] where \[F_{\mu\nu}(y) = \hat\partial_\mu A_\nu(y) - \hat\partial_\nu A_\mu(y) - i [A_\mu(y),A_\nu(y)] ~, \qquad D_\mu \bar\lambda(y) = \hat\partial_\mu \bar\lambda(y) - i [A_\mu(y),\bar\lambda(y)] ~.\]

We are now in a position to construct supersymmetric actions for the non-abelian vector superfield. The first type of term we consider is the kinetic term \[\begin{split} \mathcal{S}_{\text{YM}} & = \mathop{\mathrm{Im}}\Big(\int d^4x d^2\theta \, \frac{\tau}{4\pi} \mathop{\mathrm{Tr}}(W^\alpha W_\alpha) \Big) \\ & = \int d^4 x \, \big(\frac{1}{g^2}\mathop{\mathrm{Tr}}(-\frac{1}{2} F^{\mu\nu} F_{\mu\nu} - 2i \bar\lambda \bar\sigma^\mu D_\mu \lambda + D^2) + \frac{\theta}{16\pi^2} \mathop{\mathrm{Tr}}(\tilde{F}^{\mu\nu} F_{\mu\nu}) \big) ~, \end{split}\] where \[F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu - i [A_\mu,A_\nu] ~, \qquad D_\mu \bar\lambda = \partial_\mu \bar\lambda - i [A_\mu,\bar\lambda] ~,\] are the non-abelian field strength and the gauge covariant derivative in the adjoint representation of \(\bar\lambda\) respectively, and we use the convention that \(\mathop{\mathrm{Tr}}(T^a T^b) = \frac{1}{2} \delta^{ab}\) in the adjoint representation. Note that there is no gauge-invariant generalisation of the FI term for a non-abelian vector superfield.

The second type of term describes the coupling to matter fields \[\begin{split} \mathcal{S}_{\text{matter}} & = \int d^4x d^2\theta d^2\bar\theta \, \bar\Phi e^{2V_{\scriptscriptstyle{R}}} \Phi \\ & = \int d^4x \, \big(\bar D_{\scriptscriptstyle{R}}{}_\mu\bar\phi D_{{\scriptscriptstyle{R}}}^\mu\phi - i \bar\psi \bar\sigma^\mu D_{\scriptscriptstyle{R}}{}_\mu\psi + \bar F F + i \sqrt{2}\bar\phi \lambda_{\scriptscriptstyle{R}} \psi -i\sqrt{2}\bar\psi \bar\lambda_{\scriptscriptstyle{R}}\phi + \bar \phi D_{\scriptscriptstyle{R}} \phi\big) ~, \end{split}\] up to total derivatives. The gauge covariant derivative acting on a field transforming in the representation \(R\) is \[D_{\scriptscriptstyle{R}}{}_\mu = \partial_\mu - i A_{{\scriptscriptstyle{R}}}{}_\mu ~,\] and we have defined \[A_{{\scriptscriptstyle{R}}}{}_\mu = A_{\mu a} T_{\scriptscriptstyle{R}}^a ~, \qquad \lambda_{{\scriptscriptstyle{R}}} = \lambda_a T_{\scriptscriptstyle{R}}^a ~, \qquad D_{{\scriptscriptstyle{R}}} = D_a T_{\scriptscriptstyle{R}}^a ~.\] The gauge indices in \(\mathcal{S}_{\text{matter}}\) are contracted to extract the singlet out of \(\bar R \otimes R\), e.g. \(\bar\phi D_{\scriptscriptstyle{R}} \phi = \bar\phi_i D_a (T_{\scriptscriptstyle{R}}^a)^i{}_j \phi^j\). We may also have superpotential terms \[\begin{split} \mathcal{S}_{\text{W}} & = \int d^4x d^2\theta \, W(\Phi) + \int d^4x d^2\bar\theta \, \bar W(\bar\Phi) ~, \end{split}\] where \(W(\Phi)\) is a gauge invariant holomorphic function of \(\Phi\).

From these three types of terms we can construct non-abelian supersymmetric gauge theories. Taking the gauge group to be \(\mathrm{G} = \bigotimes_A \mathrm{G}_A\), we have a non-abelian vector superfield \(V_A\) for each simple factor \(\mathrm{G}_A\) of \(\mathrm{G}\). We also have chiral superfields \(\Phi^I\) transforming in the representation \(R_I\) of \(\mathrm{G}\). The action is given by \[\mathcal{S}= \sum_A \mathcal{S}_{\text{YM}} + \sum_I \mathcal{S}_{\text{matter}} +\mathcal{S}_{\text{W}} ~,\] and we may also have a FI term for any abelian factors of \(\mathrm{G}\). This action contains one \(\mathcal{S}_{\text{YM}}\) term for each simple factor \(\mathrm{G}_A\) of \(\mathrm{G}\), one \(\mathcal{S}_{\text{matter}}\) term for each chiral superfield \(\Phi^I\) and a single \(\mathcal{S}_{\text{W}}\) term constructed from a gauge invariant superpotential \(W(\Phi)\).

The equations of motion for the auxiliary fields (assuming that there are no FI terms) are \[\begin{split} \bar F_I & = \partial_I W(\phi) ~, \qquad F^I = \bar\partial^I \bar W(\bar\phi) ~, \\ D_A & = - g_A^2 \sum_I \bar \phi_I T^A_{{\scriptscriptstyle{R}}_I} \phi^I = - g_A^2 \mu_A(\phi,\bar\phi) ~, \end{split}\] where \(\mu_A(\phi,\bar\phi)\) are the moment maps of \(\mathrm{G}_A\), and \(T^A_{{\scriptscriptstyle{R}}_I}\) are the generators in the representation \(R_I\) of \(\mathrm{G}_A\). Integrating out the auxiliary fields we find the scalar potential \[\begin{split} V(\phi,\bar\phi) & = \sum_I \bar F_I F^I + \sum_A \frac{1}{2 g_A^2} (D_A)^2 \\ & = \sum_I \bar\partial^I\bar W(\bar \phi) \partial_I W(\phi) + \sum_A \frac{g_A^2}{2} (\mu_A(\phi,\bar\phi))^2 ~, \end{split}\] which again has the general structure of |F-term|\(^2\) + (D-term)\(^2\). The moduli space of supersymmetric vacua is then given by \[\begin{split} \mathcal{M} & = \Big\{(\phi,\bar\phi) ~:~ \bar F_I = \partial_I W(\phi) = 0 ~~\text{and}~~ D_A = -g_A^2 \mu_A(\phi,\bar\phi) = 0 \Big\}\Big/ \mathrm{G} \\ & = \Big\{ \phi ~:~ \partial_I W(\phi) = 0 \Big\} \Big/ \mathrm{G}^\mathbb{C}~, \end{split}\] where \(\mathrm{G}^\mathbb{C}\) is the complexified gauge group. Just as in the abelian case, \(\mathcal{M}\) is parametrised by the vacuum expectation value of gauge invariant chiral operators.

Let us conclude with an example of a non-abelian supersymmetric gauge theory, \(\mathrm{SU}(N_c)\) supersymmetric quantum chromodynamics (SQCD) with \(N_f\) flavours. We have a non-abelian vector superfield with \(\mathrm{SU}(N_c)\) gauge group. In addition, we have matter fields, i.e. chiral superfields, \(Q\) and \(\tilde{Q}\). The theory has a number of different symmetries:

\(\mathrm{SU}(N_c)\) gauge symmetry – \(Q\) transforms in the fundamental representation and \(\tilde Q\) in the antifundamental representation;

\(\mathrm{SU}(N_f)_{_L} \times \mathrm{SU}(N_f)_{_R}\) global symmetry – \(Q\) transforms in the antifundamental representation of \(\mathrm{SU}(N_f)_{_L}\) and the trivial representation of \(\mathrm{SU}(N_f)_{_R}\), \(\tilde Q\) transforms in the trivial representation of \(\mathrm{SU}(N_f)_{_L}\) and the fundamental representation of \(\mathrm{SU}(N_f)_{_R}\);

\(\mathrm{U}(1)_{_B} \times \mathrm{U}(1)_{_A} \times \mathrm{U}(1)_{_R}\) global symmetry – \(Q\) has charges \([Q]_{_B} = 1\), \([Q]_{_A} = 1\) and \([Q]_{_R} = 0\), \(\tilde{Q}\) has charges \([\tilde Q]_{_B} = -1\), \([\tilde Q]_{_A} = 1\) and \([\tilde Q]_{_R} = 0\).

The gauge invariant chiral operators in this theory are the mesons \[M_{\tilde\imath}{}^j = \tilde Q_{\tilde\imath}^a Q_a^j ~,\] and the baryons and antibaryons \[B^{j_1\dots j_{N_c}} = \epsilon^{a_1\dots a_{N_c}} Q_{a_1}^{j_1} \dots Q_{a_{N_c}}^{j_{N_c}} ~, \qquad \tilde B_{\tilde\imath_1\dots \tilde\imath_{N_c}} = \epsilon_{a_1\dots a_{N_c}} \tilde Q^{a_1}_{\tilde\imath_1} \dots \tilde Q^{a_{N_c}}_{\tilde\imath_{N_c}} ~,\] where \(a=1,\dots, N_c\) is the index of the \(\mathrm{SU}(N_c)\) gauge group, while \(i=1,\dots, N_f\) and \(\tilde\imath=1,\dots, N_f\) are the indices of the \(\mathrm{SU}(N_f)_{_L} \times \mathrm{SU}(N_f)_{_R}\) global symmetry group. Baryons and antibaryons can only exist if \(N_f \geq N_c\), and the \(\mathrm{U}(1)_{_B}\) charge is the baryon number: \([M_{\tilde\imath}{}^j]_{_B} = 0\), \([B^{j_1\dots j_{N_c}}]_{_B} = N_c\) and \([\tilde B_{\tilde\imath_1\dots \tilde\imath_{N_c}}]_{_B} = -N_c\).

A non-abelian vector superfield \(V\) transforms under supersymmetric gauge transformations as \[e^{2V} \to e^{i\bar\Delta} e^{2V} e^{-i\Delta} ~,\] where \(\Delta\) is a chiral superfield valued in the Lie algebra of the gauge group, so that \(e^{i\Delta}\) is a chiral superfield valued in the gauge group itself. \(\bar\Delta = \Delta^\dagger\) is the conjugate antichiral superfield.

Check that the gaugino superfield \[W_\alpha = -\frac{1}{8} \bar D^2(e^{-2V}D_\alpha e^{2V}) ~,\] transforms as \(W_\alpha \to e^{i\Delta} W_\alpha e^{-i \Delta}\) under supersymmetric gauge transformations.

Derive the component field expansion of \(W_\alpha\) in the Wess-Zumino gauge \[W_{\text{WZ}}{}_\alpha(y,\theta) = - i \lambda_\alpha(y) + \theta_\alpha D(y) + i (\sigma^{\mu\nu}\theta)_\alpha F_{\mu\nu}(y) + (\theta\theta)(\sigma^\mu D_\mu \bar \lambda(y))_\alpha ~,\] where \[F_{\mu\nu}(y) = \hat\partial_\mu A_\nu(y) - \hat\partial_\nu A_\mu(y) - i [A_\mu(y),A_\nu(y)] ~, \qquad D_\mu \bar\lambda(y) = \hat\partial_\mu \bar\lambda(y) - i [A_\mu(y),\bar\lambda(y)] ~,\] are the non-abelian field strength and the gauge covariant derivative in the adjoint representation of \(\bar\lambda\) respectively. Recall that \(\hat\partial_\mu\) just denotes the partial derivative with respect to \(y^\mu\).

Under supersymmetric gauge transformations \[\begin{split} W_\alpha & \to -\frac{1}{8}\bar D^2(e^{i\Delta} e^{-2V} e^{-i\bar\Delta} D_\alpha (e^{i\bar\Delta} e^{2V} e^{-i \Delta})) = -\frac{1}{8} e^{i\Delta} \bar D^2 (e^{-2V} D_\alpha(e^{2V} e^{-i\Delta})) \\ & \qquad = -\frac{1}{8} e^{i\Delta} \bar D^2(e^{-2V}(D_\alpha e^{2V})e^{-i\Delta}) = -\frac{1}{8} e^{i\Delta} \bar D^2(e^{-2V}(D_\alpha e^{2V}))e^{-i\Delta} \\ & \qquad = e^{i\Delta} W_\alpha e^{-i\Delta} ~, \end{split}\] where we have used that \(\bar D_{\dot\alpha} e^{\pm i\Delta} = D_\alpha e^{\pm i\bar\Delta} = 0\) and \[\bar D^2 D_\alpha e^{-i\Delta} = -\bar D^{\dot\alpha} \bar D_{\dot\alpha} D_\alpha e^{-i\Delta} = -\bar D^{\dot\alpha}\{\bar D_{\dot\alpha}, D_\alpha\} e^{-i\Delta} = -2i (\sigma^\mu)_{\alpha\dot\alpha} \bar D^{\dot\alpha} \partial_\mu e^{-i\Delta} = 0 ~,\] since \(\Delta\) and \(\bar\Delta\) are chiral and antichiral superfields respectively. Therefore, we see that \(W_\alpha\) is gauge covariant.

To derive the component field expansion of \(W_\alpha\) in the Wess-Zumino gauge, we start by expanding out the exponentials \[\begin{split} W_{\text{WZ}}{}_\alpha & = - \frac{1}{8} \bar D^2 ((1-2 V_{\text{WZ}} + 2 V_{\text{WZ}}^2)D_\alpha(1+2 V_{\text{WZ}} + 2 V_{\text{WZ}}^2)) ~, \\ & =- \frac{1}{4} \bar D^2(D_\alpha V_{\text{WZ}} -2 V_{\text{WZ}} D_\alpha V_{\text{WZ}} + D_\alpha (V_{\text{WZ}}^2)) \\ & =- \frac{1}{4} \bar D^2(D_\alpha V_{\text{WZ}} +[ D_\alpha V_{\text{WZ}},V_{\text{WZ}}]) \end{split}\] where we have used that \(V_{\text{WZ}}^3 = (D_\alpha V_{\text{WZ}}) V_{\text{WZ}}^2 = V_{\text{WZ}} (D_\alpha V_{\text{WZ}}) V_{\text{WZ}} = V_{\text{WZ}}^2 (D_\alpha V_{\text{WZ}}) = 0\). In the Wess-Zumino gauge \[V_{\text{WZ}}(x,\theta,\bar\theta) = (\theta\sigma^\mu\bar\theta) A_\mu(x) + i (\theta\theta)(\bar\theta\bar\lambda(x)) - i (\bar\theta\bar\theta)(\theta\lambda(x)) + \frac{1}{2}(\theta\theta)(\bar\theta\bar\theta) D(x) ~.\] In terms of the coordinates \((y,\theta,\bar\theta)\) we have \(D_\alpha = \hat\partial_\alpha + 2i(\sigma^\mu\bar\theta)_\alpha \hat\partial_\mu\), \(\bar D_{\dot\alpha} = \hat{\bar\partial}_{\dot\alpha}\) and \[\begin{split} V_{\text{WZ}}(y,\theta,\bar\theta) & = (\theta\sigma^\mu\bar\theta) \big(A_\mu(y)-i (\theta\sigma^\nu\bar\theta)\hat\partial_\nu A_\mu\big) + i (\theta\theta) (\bar\theta\bar\lambda(y)) - i (\bar\theta\bar\theta)(\theta\lambda(y)) + \frac{1}{2} (\theta\theta) (\bar\theta\bar\theta) D(y) \\ & = (\theta\sigma^\mu\bar\theta) A_\mu(y) + i (\theta\theta) (\bar\theta\bar\lambda(y)) - i (\bar\theta\bar\theta)(\theta\lambda(y)) + \frac{1}{2} (\theta\theta) (\bar\theta\bar\theta) \big(D(y)-i \hat\partial_\mu A^\mu(y)\big) ~. \end{split}\] Therefore, \[\begin{split} D_\alpha V_{\text{WZ}}(y,\theta,\bar\theta) & = (\sigma^\mu\bar\theta)_\alpha A_\mu(y) + 2i (\sigma^\nu\bar\theta)_\alpha(\theta\sigma^\mu\bar\theta) \hat\partial_\nu A_\mu(y) + 2i \theta_\alpha (\bar\theta\bar\lambda(y)) - 2 (\sigma^\mu\bar\theta)_\alpha (\theta\theta) (\bar\theta\hat\partial_\mu \bar\lambda(y)) \\ & \quad - i (\bar\theta\bar\theta)\lambda_\alpha(y) + \theta_\alpha (\bar\theta\bar\theta) \big(D(y)-i \hat\partial_\mu A^\mu(y)\big) \\ & = (\sigma^\mu\bar\theta)_\alpha A_\mu(y) + i (\sigma^\nu\bar\sigma^\mu\theta)_{\alpha} (\bar\theta\bar\theta) \hat\partial_\nu A_\mu(y) -i \theta_\alpha (\bar\theta\bar\theta) \hat\partial_\mu A^\mu(y) \\ & \quad + 2i \theta_\alpha (\bar\theta\bar\lambda(y)) + (\theta\theta)(\bar\theta\bar\theta) (\sigma^\mu\hat\partial_\mu\bar\lambda(y))_{\alpha} - i (\bar\theta\bar\theta)\lambda_\alpha(y) + \theta_\alpha (\bar\theta\bar\theta) D(y) \\ & = (\sigma^\mu\bar\theta)_\alpha A_\mu(y) + \frac{i}{2} ((2\eta^{\mu\nu}\mathbf{1}- 4\sigma^{\mu\nu})\theta)_{\alpha} (\bar\theta\bar\theta) \hat\partial_\nu A_\mu(y) -i \theta_\alpha (\bar\theta\bar\theta) \hat\partial_\mu A^\mu(y) \\ & \quad + 2i \theta_\alpha (\bar\theta\bar\lambda(y)) + (\theta\theta)(\bar\theta\bar\theta) (\sigma^\mu\hat\partial_\mu\bar\lambda(y))_{\alpha} - i (\bar\theta\bar\theta)\lambda_\alpha(y) + \theta_\alpha (\bar\theta\bar\theta) D(y) \\ & = (\sigma^\mu\bar\theta)_\alpha A_\mu(y) + i (\sigma^{\mu\nu}\theta)_{\alpha} (\bar\theta\bar\theta) (\hat\partial_\mu A_\nu(y) - \hat\partial_\nu A_\mu(y)) \\ & \quad + 2i \theta_\alpha (\bar\theta\bar\lambda(y)) + (\theta\theta)(\bar\theta\bar\theta) (\sigma^\mu\hat\partial_\mu\bar\lambda(y))_{\alpha} - i (\bar\theta\bar\theta)\lambda_\alpha(y) + \theta_\alpha (\bar\theta\bar\theta) D(y) ~, \end{split}\] where we recall that \[\begin{aligned} \sigma^\mu\bar\sigma^\nu - \sigma^\nu\bar\sigma^\mu & = 4\sigma^{\mu\nu} ~, \qquad & \bar\sigma^\mu\sigma^\nu - \bar\sigma^\nu\sigma^\mu & = 4\bar\sigma^{\mu\nu} ~, \\ \sigma^\mu\bar\sigma^\nu + \sigma^\nu\bar\sigma^\mu & = 2\eta^{\mu\nu}\mathbf{1}~, \qquad & \bar\sigma^\mu\sigma^\nu + \bar\sigma^\nu\sigma^\mu & = 2\eta^{\mu\nu}\mathbf{1}~. \end{aligned}\] We also have \[\begin{split} [D_\alpha V_{\text{WZ}},V_{\text{WZ}}] (y,\theta,\bar\theta) & = [(\sigma^\mu\bar\theta)_\alpha A_\mu(y)+ 2i \theta_\alpha (\bar\theta\bar\lambda(y)),(\theta\sigma^\nu\bar\theta) A_\nu(y) + i (\theta\theta)(\bar\theta\bar\lambda(y))] \\ & = (\sigma^\mu\bar\theta)_\alpha(\theta\sigma^\nu\bar\theta) [A_\mu(y),A_\nu(y)] \\ & \quad + i (\sigma^\mu\bar\theta)_\alpha (\theta\theta) [A_\mu(y),(\bar\theta\bar\lambda(y))] + 2i (\theta\sigma^\nu\bar\theta)\theta_\alpha[ (\bar\theta\bar\lambda(y)), A_\nu(y) ] \\ & = \frac{1}{2}(\sigma^\mu\bar\sigma^\nu\theta)_\alpha (\bar\theta\bar\theta) [A_\mu(y),A_\nu(y)] \\ & \quad + i (\sigma^\mu\bar\theta)_\alpha (\theta\theta) [A_\mu(y),(\bar\theta\bar\lambda(y))] - i (\sigma^\nu\bar\theta)_\alpha(\theta\theta)[ (\bar\theta\bar\lambda(y)), A_\nu(y) ] \\ & = (\sigma^{\mu\nu}\theta)_\alpha (\bar\theta\bar\theta) [A_\mu(y),A_\nu(y)] - i (\theta\theta)(\bar\theta\bar\theta)[A_\mu(y), (\sigma^\mu\bar\lambda(y))_\alpha] ~. \end{split}\] Now using that \(\hat{\bar\partial}^2(\bar\theta\bar\theta) = -4\) we find \[\begin{split} W_\alpha(y,\theta,\bar\theta) & = -\frac{1}{4} \bar D^2 (D_\alpha V_{\text{WZ}}+[D_\alpha V_{\text{WZ}}, V_{\text{WZ}}])(y,\theta,\bar\theta) \\ & = - i \lambda_\alpha(y) + i (\sigma^{\mu\nu}\theta)_{\alpha} (\hat\partial_\mu A_\nu(y) - \hat\partial_\nu A_\mu(y)) + \theta_\alpha D(y) + (\theta\theta) (\sigma^\mu\hat\partial_\mu\bar\lambda(y))_{\alpha} ~. \\ & \quad + (\sigma^{\mu\nu}\theta)_\alpha [A_\mu(y),A_\nu(y)] - i (\theta\theta)[A_\mu(y), (\sigma^\mu\bar\lambda(y))_\alpha] \\ & = - i \lambda_\alpha(y) + i (\sigma^{\mu\nu}\theta)_{\alpha} F_{\mu\nu}(y) + \theta_\alpha D(y) + (\theta\theta) (\sigma^\mu D_\mu\bar\lambda(y))_{\alpha} ~, \end{split}\] where \(F_{\mu\nu}(y) = \hat\partial_\mu A_\nu(y) - \hat\partial_\nu A_\mu(y)- i [A_\mu(y),A_\nu(y)]\) is the non-abelian field strength and \(D_\mu\bar\lambda(y) = \hat\partial_\mu \bar\lambda(y) - i [A_\mu(y),\bar\lambda(y)]\) is the gauge covariant derivative in the adjoint representation of \(\bar\lambda\).

The minimal supersymmetric Standard Model (MSSM) is the minimal supersymmetric extension of the Standard Model. Its field content is obtained by promoting the field content of the Standard Model to superfields as follows:

The gauge fields \(\longrightarrow\) vector superfields;

The left-handed fermions and the conjugates of the right-handed fermions \(\longrightarrow\) chiral superfields;

The Higgs field \(\longrightarrow\) 2 chiral superfields \(H_u\) and \(H_d\).

In the MSSM the Higgs field is promoted to two chiral superfields in order to cancel gauge anomalies and to ensure that the Yukawa couplings of the Standard Model follow from a superpotential. Under the \(\mathrm{SU}(3)_{_c} \times \mathrm{SU}(2)_{_L} \times \mathrm{U}(1)_{_Y}\) gauge symmetry the chiral superfields transform in the following representations

\(\mathrm{SU}(3)_{_c}\) | \(\mathrm{SU}(2)_{_L}\) | \(\mathrm{U}(1)_{_Y}\) | |
---|---|---|---|

\(Q_i\) | \(\mathbf{3}\) | \(\mathbf{2}\) | \(\frac{1}{6}\) |

\(U_i^\dagger\) | \(\bar{\mathbf{3}}\) | \(\mathbf{1}\) | \(-\frac{2}{3}\) |

\(D_i^\dagger\) | \(\bar{\mathbf{3}}\) | \(\mathbf{1}\) | \(\frac{1}{3}\) |

\(L_i\) | \(\mathbf{1}\) | \(\mathbf{2}\) | \(-\frac{1}{2}\) |

\(E_i^\dagger\) | \(\mathbf{1}\) | \(\mathbf{1}\) | \(1\) |

\(H_u\) | \(\mathbf{1}\) | \(\mathbf{2}\) | \(\frac{1}{2}\) |

\(H_d\) | \(\mathbf{1}\) | \(\mathbf{2}\) | \(-\frac{1}{2}\) |

where the index \(i=1,2,3\) labels the three generations of the Standard Model. The superpotential is given by \[W = \mu H_u H_d + (y_u)_{ij} H_u Q_i U_j^\dagger + (y_d)_{ij} H_d Q_i D_j^\dagger + (y_l)_{ij} H_d L_i E_j^\dagger ~.\] This superpotential is renormalisable, gauge invariant and preserves R-parity \(P_R = (-1)^{3(B-L) + 2s}\), where \(B\) is the baryon number, \(L\) is the lepton number and \(s\) is the spin. \(P_R\) is \(+1\) for the Standard Model particles and \(-1\) for their superpartners. Demanding invariance under R-parity forbids those couplings in the superpotential that would lead to couplings not present in the Standard Model.

If supersymmetry is a symmetry of the laws of nature, then it must be broken spontaneously since the number of bosonic degrees of freedom in the Standard Model is different to the number of fermionic degrees of freedom. Moreover, we do not observe any mass degeneracy between bosons and fermions with the same charges. Therefore, supersymmetry must be broken spontaneously at energies below some energy scale \(M_{\text{SUSY}}\) and restored at energies above \(M_{\text{SUSY}}\) through threshold corrections with \[1~\text{TeV} \quad \lessapprox \quad M_{\text{SUSY}} \quad \lessapprox \quad M_{\text{GUT}} ~\text{or}~ M_{\text{P}} ~.\] The lower bound of \(1~\text{TeV}\) comes from the fact that we do not observe any superpartners at this energy scale, and we recall that \(M_{\text{GUT}} \sim 10^{15}/10^{16}~\text{GeV}\) and \(M_{\text{P}} \sim 10^{19}~\text{GeV}\). It is often assumed that \(M_{\text{SUSY}}\) is close to \(1~\text{TeV}\) in order to maintain and possibly explain the hierarchy between \(M_{\text{EW}} \sim 250~\text{GeV}\) and \(M_{\text{GUT}}\) or \(M_{\text{P}}\). However, it may be the case that supersymmetry is only restored at a higher energy scale.

Let us now compare spontaneous supersymmetry breaking to the spontaneous breaking of internal global or gauge symmetries. The spontaneous breaking of internal global or gauge symmetries is controlled by the locations of the minima of the scalar potential. Spontaneous supersymmetry breaking is controlled by the minimum of the potential itself.

In this section, since we will be interested in the physical mass spectrum, we rescale vector multiplets \(V \to g V\) so that the kinetic terms for the gauge field are canonically normalised \[\mathcal{S}_{\text{YM}} = \int d^4x \, \big(\mathop{\mathrm{Tr}}(-\frac{1}{2} F^{\mu\nu}F_{\mu\nu}) + \dots\big) ~, \qquad F_{\mu\nu} = \partial_\mu A_{\nu} - \partial_\nu A_{\mu} - i g[A_\mu,A_\nu] ~.\] Taking the gauge group to be \(\mathrm{G} = \bigotimes_A \mathrm{G}_A\), the scalar potential is then given by \[\begin{split} V(\phi,\bar\phi) & = \sum_I \bar F_I F^I + \frac{1}{2} \sum_A (D_A)^2 \\ & = \sum_I \bar\partial^I \bar W(\bar\phi) \partial_I W(\phi) + \sum_A \frac{g_A^2}{2}(\sum_I \bar \phi_I T^A_{{\scriptscriptstyle{R}}_I} \phi^I - \xi_A)^2 ~, \end{split}\] where \(\xi_A \neq 0\) only if \(\mathrm{G}_A = \mathrm{U}(1)\). We then have that supersymmetry is broken spontaneously if and only if \(V_{\text{min}} > 0\), which follows from the \(\mathcal{N}=1\) super-Poincaré algebra. From the structure of the scalar potential we see that this is satisfied if and only if \(\langle D_A\rangle \neq 0\) or \(\langle F^I\rangle \neq 0\). This is consistent with the supersymmetry transformations of the component fields since for Poincaré invariant vacua only scalar fields can acquire a non-zero vacuum expectation value. Therefore, \[\begin{aligned} \delta_{\varepsilon,\bar\varepsilon} \langle \phi^I \rangle & = 0 ~, & \qquad \delta_{\varepsilon,\bar\varepsilon} \langle \psi^I \rangle & \sim \varepsilon \langle F^I \rangle ~, & \qquad \delta_{\varepsilon,\bar\varepsilon} \langle F^I \rangle & = 0 ~, \\ \delta_{\varepsilon,\bar\varepsilon} \langle A_{A\mu} \rangle & = 0 ~, & \qquad \delta_{\varepsilon,\bar\varepsilon} \langle \lambda_A \rangle & \sim \varepsilon \langle D_A \rangle ~, & \qquad \delta_{\varepsilon,\bar\varepsilon} \langle D_A \rangle & = 0 ~, \end{aligned}\] and the vacuum indeed spontaneously breaks supersymmetry if and only if \(\langle D_A\rangle \neq 0\) or \(\langle F^I\rangle \neq 0\).

The Goldstone theorem states that when a continuous global symmetry is spontaneously broken, there is a massless mode in the spectrum whose quantum numbers match those of the broken symmetry generator. This assumes that Poincaré symmetry of Minkowski space-time is not broken. If supersymmetry is spontaneously broken then we expect a massless Weyl fermion, which is often called the goldstino. The presence of the goldstino can be shown using non-perturbative current algebra techniques only using supersymmetry, but here we will give a more direct derivation based on the classical action.

Suppose that there is a vacuum that spontaneously breaks supersymmetry \[\frac{\partial V}{\partial \phi^I} = 0 \qquad \Rightarrow \qquad \sum_J \langle \partial_I \partial_J W \rangle \langle F^J \rangle - \sum_A g_A \langle \bar\phi_I \rangle T^A_{{\scriptscriptstyle{R}}_I} \langle D_A \rangle = 0 ~,\] with either \(\langle F^I\rangle \neq 0\) or \(\langle D_A \rangle \neq 0\). The gauge invariance of the superpotential implies that \[\sum_I \langle \partial_I W \rangle T^A_{{\scriptscriptstyle{R}}_I} \langle \phi^I \rangle = \sum_I \langle \bar F_I \rangle T^A_{{\scriptscriptstyle{R}}_I} \langle \phi^I \rangle = 0 \qquad \Rightarrow \qquad \sum_I \langle \bar \phi_I \rangle T^A_{{\scriptscriptstyle{R}}_I} \langle F^I \rangle = 0 ~.\] These equations can be combined into the matrix equation \[\sum_J \sum_A \tilde M_{\frac{1}{2}} \begin{pmatrix} \langle F^J \rangle \\ \langle D_A \rangle \end{pmatrix} = 0 ~, \qquad \tilde M_{\frac{1}{2}} = \begin{pmatrix} \langle \partial_I \partial_J W \rangle & - g_A \langle \bar\phi_I \rangle T^A_{{\scriptscriptstyle{R}}_I} \\ -g_B \langle \bar\phi_J \rangle T^B_{{\scriptscriptstyle{R}}_J} & 0 \end{pmatrix} ~.\] Since supersymmetry is broken spontaneously, the vector \(\big(\begin{smallmatrix} \langle F^J \rangle \\ \langle D_A \rangle \end{smallmatrix}\big) \neq 0\) is a zero eigenvector of \(\tilde M_{\frac{1}{2}}\).

Comparing with the action, we see that \(\tilde M_{\frac{1}{2}}\) is closely related to the complex mass matrix of the fermions \[\mathcal{S}= \int d^4x \, \big( \dots -\frac{1}{2} \sum_{I,J}\sum_{A,B}\begin{pmatrix} \psi^I &\lambda_B \end{pmatrix} M_{\frac{1}{2}} \begin{pmatrix} \psi^J \\ \lambda_A \end{pmatrix} + \dots \big) ~, \qquad M_{\frac{1}{2}} = \begin{pmatrix}1 & 0 \\ 0 & \sqrt{2}i \end{pmatrix} \tilde M_{\frac{1}{2}}\begin{pmatrix}1 & 0 \\ 0 & \sqrt{2}i \end{pmatrix} ~.\] It therefore follows that, expanding around a vacuum that spontaneously breaks supersymmetry, the spectrum contains a massless fermion, the goldstino \[\begin{pmatrix} \psi^J \\ \sqrt{2}i \lambda_A \end{pmatrix} \propto \begin{pmatrix} \langle F^J \rangle \\ \langle D_A \rangle \end{pmatrix} \psi_{\text{G}} ~.\]

There are three possible ways that supersymmetry can be broken spontaneously:

F-term spontaneous supersymmetry breaking: \(\langle F^I\rangle \neq 0\), \(\langle D_A \rangle = 0\), \(\psi_{\text{G}}\) is a linear combination of the matter fermions \(\psi^I\).

D-term spontaneous supersymmetry breaking: \(\langle F^I\rangle = 0\), \(\langle D_A \rangle \neq 0\), \(\psi_{\text{G}}\) is a linear combination of the gauginos \(\lambda_A\).

Mixed F-term and D-term spontaneous supersymmetry breaking: \(\langle F^I\rangle \neq 0\), \(\langle D_A \rangle \neq 0\), \(\psi_{\text{G}}\) is a linear combination of the matter fermions and gauginos.

**F-term spontaneous supersymmetry breaking and the O’Raifeartaigh model.**

In order to have F-term spontaneous supersymmetry breaking we need a superpotential with linear terms. Otherwise setting all chiral superfields equal to zero will solve \(F^I = 0\). The O’Raifeartaigh model is a Wess-Zumino model with superpotential \[W(X,Y,Z) = Y(a^2 - X^2) + b XZ ~,\] where \(a\) and \(b\) are non-zero complex parameters. For this superpotential we have \[\bar F_X = - 2\phi^X \phi^Y +b \phi^Z ~, \qquad \bar F_Y = a^2 - (\phi^X)^2 ~, \qquad \bar F_Z = b \phi^X ~.\] It is straightforward to see that we cannot set \(\bar F_Y\) and \(\bar F_Z\) to zero simultaneously, hence supersymmetry is broken spontaneously.

The O’Raifeartaigh model is a Wess-Zumino model of three chiral superfields with superpotential \[W(X,Y,Z) = Y(a^2 - X^2) + b XZ ~,\] where \(a\) and \(b\) are non-zero complex parameters. Show that supersymmetry is spontaneously broken. Assuming \(2|a|^2<|b|^2\), find the global minimum (or minima) of the scalar potential and determine the vacuum expectation values of the scalar potential.

For the superpotential \[W(X,Y,Z) = Y(a^2 - X^2) + b XZ ~,\] we have \[\begin{split} \bar F_X = \frac{\partial W}{\partial X}\Big|_{X = \phi^X,Y=\phi^Y,Z=\phi^Z} & = - 2 \phi^X \phi^Y +b \phi^Z ~, \\ \bar F_Y = \frac{\partial W}{\partial Y}\Big|_{X = \phi^X,Y=\phi^Y,Z=\phi^Z} & = a^2 - (\phi^X)^2 ~, \\ \bar F_Z = \frac{\partial W}{\partial Z}\Big|_{X = \phi^X,Y=\phi^Y,Z=\phi^Z} & = b \phi^X ~. \end{split}\] \(\bar F_Z = 0\) implies that \(\phi^X = 0\). However, this implies that \(\bar F_Y = a^2 \neq 0\). Therefore, supersymmetry is spontaneously broken.

The scalar potential is given by \[\begin{split} V(\phi^X,\phi^Y,\phi^Z,\bar\phi_X,\bar\phi_Y,\bar\phi_Z) & = \bar F_X F^X + \bar F_Y F^Y + \bar F_Z F^Z ~. \end{split}\] The stationary points of the potential solve \[\begin{split} \frac{\partial V}{\partial \phi^X} & = -2 \phi^Y(-2 \bar\phi_X\bar\phi_Y + \bar b \bar\phi_Z) - 2 \phi^X(\bar a^2 - (\bar\phi_X)^2) + b \bar b \bar\phi_X = 0 ~, \\ \frac{\partial V}{\partial \phi^Y} & = -2 \phi^X(-2 \bar\phi_X\bar\phi_Y + \bar b \bar\phi_Z) = 0~, \\ \frac{\partial V}{\partial \phi^Z} & = b(-2 \bar\phi_X\bar\phi_Y + \bar b \bar\phi_Z) = 0~. \end{split}\] From the third equation we find that \[\phi^Z = \frac{2\phi^X \phi^Y}{b} ~.\] The second equation is then solved, while the first equation becomes \[2 \bar a^2 \phi^X = (2 |\phi^X|^2 + |b|^2) \bar\phi_X ~.\] This implies \[2 |a|^2 |\phi^X| = (2 |\phi^X|^2 + |b|^2 ) |\phi^X| \qquad \Rightarrow \qquad (2 |\phi^X|^2 + |b|^2 -2 |a|^2) |\phi^X| = 0~.\] Given \(|\phi^X|^2 > 0\) and we assume that \(2|a|^2 < |b|^2\), the only solution to this equation is \(|\phi^X| =0\), which implies that \(\phi^X = 0\). It follows that the stationary points of the potential are \[\langle\phi^X\rangle = \langle\phi^Z\rangle = 0 ~, \qquad \langle\phi^Y\rangle ~\text{arbitrary} ~.\] Therefore, \(\langle\bar F_X\rangle = \langle\bar F_Z\rangle = 0\) and \(\langle\bar F_Y\rangle = a^2\), hence the vacuum expectation value of the scalar potential is \[\langle V \rangle = |a|^4 ~.\] Since the scalar potential is bounded below and \(\langle V\rangle\) takes the same value at all the stationary points, it follows that the stationary points are minima, i.e. stable vacua. Note that \(\phi^Y\) is a flat direction of the potential.

The Polonyi model is a Wess-Zumino model of a single chiral superfield \(X\) with linear superpotential \[W(X) = f X ~,\] where \(f\) is a non-zero complex parameter. The Polonyi model can be modified by adding a small cubic term to the superpotential \[W_\epsilon (X) = f X + \epsilon \frac{\lambda}{3} X^3 ~, \qquad 0 < \epsilon \ll 1 ~,\] where \(\lambda\) is a non-zero complex parameter. Show that supersymmetry is broken spontaneously in the Polonyi model and compute the mass spectrum of bosons and fermions. Show that supersymmetry is restored in the modified Polonyi model and that \(\langle \phi^X \rangle = 0\) is an unstable vacuum, i.e. a stationary point of the potential that is not a minimum. What happens to the supersymmetric vacua and to the scalar potential of the modified Polonyi model in the limit \(\epsilon \to 0\)?

For the superpotential of the Polonyi model \[W(X) = f X ~,\] we have \[\bar F_X = \frac{\partial W}{\partial X}\Big|_{X = \phi^X} = f ~.\] Since \(\bar F_X \neq 0\) supersymmetry is spontaneously broken. The action of the Wess-Zumino model in terms of component fields is given by \[\begin{split} \mathcal{S}& = \int d^4x \, \big(\partial_\mu\bar\phi_X \partial^\mu\phi^X - i\bar\psi_X \bar\sigma^\mu \partial_\mu \psi^X \\ & \hspace{50pt} - W'(\phi^X) \bar W'(\bar\phi_X) -\frac{1}{2}W''(\phi^X)\psi^X\psi^X -\frac{1}{2}\bar W''(\bar\phi_X)\bar\psi_X\bar\psi_X \big) ~. \end{split}\] Using that \(W'(\phi^X) = f\) and \(W''(\phi^X) = 0\) the action becomes \[\begin{split} \mathcal{S}& = \int d^4x \, \big(\partial_\mu\bar\phi_X \partial^\mu\phi^X - i\bar\psi_X \bar\sigma^\mu \partial_\mu \psi^X - f\bar f \big) ~. \end{split}\] Therefore, we can straightforwardly read off that the complex scalar and Majorana fermion are both massless.

For the superpotential of the modified Polonyi model \[W_\epsilon (X) = f X + \epsilon \frac{\lambda}{3} X^3 ~, \qquad 0 < \epsilon \ll 1 ~,\] we have \[\bar F_X = \frac{\partial W}{\partial X}\Big|_{X = \phi^X} = f + \epsilon \lambda (\phi^X)^2 ~.\] Therefore, we can solve \(\bar F_X = 0\) and we have the supersymmetric vacuum \[\langle \phi^X \rangle = \sqrt{-\frac{f}{\epsilon\lambda}} ~.\] The scalar potential is given by \[V(\phi^X,\bar\phi_X) = \bar F_X F^X = (f + \epsilon \lambda (\phi^X)^2 )(\bar f + \epsilon \bar \lambda (\bar \phi_X)^2 ) ~.\] The stationary points of the potential solve \[\frac{\partial V}{\partial \phi^X} = 2\epsilon \lambda \phi^X(\bar f + \epsilon \bar \lambda (\bar \phi_X)^2 ) = 0~.\] In addition to the supersymmetric vacuum, we have the additional stationary point \[\langle \phi^X \rangle = 0 ~.\] Setting \(\phi^X = \frac{1}{\sqrt{2}} (\varphi_r^X + i \varphi_i^X)\), the Hessian matrix at this stationary point is \[\begin{pmatrix} \frac{\partial^2 V}{\partial \varphi_r^X\partial \varphi_r^X } & \frac{\partial^2 V}{\partial \varphi_r^X\partial \varphi_i^X } \\ \frac{\partial^2 V}{\partial \varphi_i^X\partial \varphi_r^X } & \frac{\partial^2 V}{\partial \varphi_i^X\partial \varphi_i^X } \end{pmatrix} \Bigg|_{ \varphi_r^X = \varphi_i^X = 0} = \begin{pmatrix} \epsilon(\bar f\lambda+f\bar\lambda) & i\epsilon(\bar f\lambda-f\bar\lambda) \\ i\epsilon(\bar f\lambda-f\bar\lambda) & -\epsilon(\bar f\lambda+f\bar\lambda) \end{pmatrix} ~,\] which has eigenvalues \(\pm 2 \epsilon |f| |\lambda|\). Since one of these eigenvalues is negative it follows that the vacuum \(\langle \phi^X \rangle = 0\) is unstable, i.e. it is not a minimum of the potential. As \(\epsilon \to 0\), the supersymmetric vacuum \(\langle \phi^X \rangle = \sqrt{-\frac{f}{\epsilon\lambda}} \to \infty\sqrt{-\frac{f}{\lambda}}\), while the scalar potential becomes a constant \[V(\phi^X,\bar\phi_X) = (f + \epsilon \lambda (\phi^X)^2 )(\bar f + \epsilon \bar \lambda (\bar \phi_X)^2 ) \to f \bar f ~.\]

**D-term spontaneous supersymmetry breaking and the Fayet-Iliopoulos mechanism**

As an example of D-term spontaneous supersymmetry breaking, consider massive SQED with matter fields \(Q\) of charge 1 and \(\tilde{Q}\) of charge \(-1\), and a non-zero FI term. Recalling that the superpotential of this model is \[W = m \tilde{Q} Q ~,\] we find \[\bar F_Q = m \phi^{\tilde{Q}} ~, \qquad \bar F_{\tilde{Q}} = m \phi^Q ~, \qquad D = -g^2 (\bar\phi_Q \phi^Q - \bar\phi_{\tilde{Q}} \phi^{\tilde{Q}} - \xi) ~.\] If \(m \neq 0\) and \(\xi \neq 0\) then we cannot set \(\bar F_Q\), \(\bar F_{\tilde{Q}}\) and \(D\) to zero at the same time.

Consider massive SQED with matter fields \(Q\) of charge 1 and \(\tilde{Q}\) of charge \(-1\), and a non-zero FI term. Assume that the complex mass \(m \neq 0\) and that the FI parameter \(\xi > 0\). Show that \(F^Q\), \(F^{\tilde{Q}}\) and \(D\) cannot be set to zero at the same time and explain why this implies that supersymmetry is broken spontaneously.

Let \(X = \frac{g^2\xi}{|m|^2}\). Find the global minimum of the scalar potential and determine the vacuum expectation values \(\langle Q\rangle\) and \(\langle\tilde{Q}\rangle\), up to gauge freedom, as functions of \(m\), \(g\) and \(X\). For \(X \leq 1\) and \(X > 1\) determine whether D-term or F-term spontaneous supersymmetry breaking occurs, or both.

Calculate the mass spectra of bosons and fermions and identify the Goldstino, i.e. the massless Goldstone fermion associated to spontaneously broken supersymmetry, in the two regimes \(X \leq 1\) and \(X > 1\).

The action of massive QED with matter fields \(Q\) of charge 1 and \(\tilde{Q}\) of charge \(-1\), and a non-zero FI term is \[\begin{split} \mathcal{S}& = \mathop{\mathrm{Im}}\Big(\int d^4 x d^2\theta \, \frac{\tau g^2}{8\pi} W^\alpha W_\alpha \Big) + \int d^4x d^2\theta d^2\bar\theta \, \big(\bar Q e^{2gV} Q + \bar{\tilde{Q}} e^{-2gV} \tilde{Q} - 2g \xi V\big) \\ & \quad + \int d^4 x d^2\theta \, m \tilde{Q} Q + \int d^4x d^2\bar\theta \, \bar m \bar{\tilde{Q}} \bar{Q} ~, \end{split}\] where \(m \neq 0\) is the complex mass parameter, \(\tau = \frac{4\pi i}{g^2} + \frac{\theta}{2\pi}\) is the complexified gauge coupling and \(\xi > 0\) is the real Fayet-Iliopoulos (FI) parameter.

The component field expansion of the chiral superfields \(Q\) and \(\tilde{Q}\) is \[Q = \phi^Q + \sqrt{2} \theta\psi^Q - \theta\theta F^Q ~, \qquad \tilde Q = \phi^{\tilde Q} + \sqrt{2} \theta\psi^{\tilde Q} - \theta\theta F^{\tilde Q} ~,\] The scalar potential is then given by \[V(\phi^Q, \phi^{\tilde{Q}}, \bar\phi_Q, \bar\phi_{\tilde{Q}}) = \bar F_Q F^Q + \bar F_{\tilde{Q}} F^{\tilde{Q}} + \frac{1}{2} D^2 ~,\] where \[\begin{aligned} F^Q & = \partial_{\bar {\phi}_Q}(\bar m \bar\phi_{\tilde{Q}} \bar\phi_Q) = \bar m \bar\phi_{\tilde{Q}} ~, \qquad & \bar F_Q & = \partial_{\phi^Q}(m \phi^{\tilde Q}\phi^{Q}) = m \phi^{\tilde{Q}} ~, \\ F^{\tilde{Q}} & = \partial_{\bar{\phi}_{\tilde{Q}}}(\bar m \bar\phi_{\tilde{Q}} \bar\phi_Q) = \bar m \bar\phi_{Q} ~, \qquad & \bar F_{\tilde{Q}} & = \partial_{\phi^{\tilde{Q}}}(m \phi^{\tilde Q} \phi^{Q}) = m \phi^Q ~, \\ D & = -g (\bar \phi_Q \phi_Q - \bar\phi_{\tilde{Q}} \phi_{\tilde Q} - \xi) ~. && \end{aligned}\] Therefore, \[V(\phi^Q, \phi^{\tilde{Q}}, \bar\phi_Q, \bar\phi_{\tilde{Q}}) = |m|^2(|\phi^Q|^2 + |\phi^{\tilde Q}|^2) + \frac{g^2}{2}(|\phi^Q|^2 - |\phi^{\tilde Q}|^2 - \xi)^2 ~.\]

To determine the moduli space of supersymmetric vacua we solve \(F^Q = F^{\tilde{Q}} = D = 0\) and quotient out by the \(\mathrm{U}(1)\) gauge symmetry \[\begin{aligned} Q & \to e^{i\alpha(x)} Q ~, \qquad & \bar Q & \to e^{-i\alpha(x)} \bar Q ~, \\ \tilde{Q} & \to e^{-i\alpha(x)} \tilde{Q} ~, \qquad & \bar{\tilde{Q}} & \to e^{i\alpha(x)} \bar{\tilde{Q}} ~. \end{aligned}\] For \(m \neq 0\) and \(\xi > 0\), we have \(F^Q = \bar m \bar \phi_{\tilde{Q}}\), \(F^{\tilde{Q}} = \bar m \bar \phi_Q\) and \(D = -g (|\phi^Q|^2 - |\phi^{\tilde Q}|^2 -\xi)\). Solving \(F^Q = F^{\tilde{Q}} = 0\) gives \(\phi^Q = \phi^{\tilde{Q}} = 0\), for which \(D = g\xi \neq 0\). Therefore, there are no supersymmetric vacua and supersymmetry is broken spontaneously. This follows since the vacuum energy is non-zero, while for supersymmetric vacua it should vanish. Furthermore, the supersymmetry transformations of the vacuum expectation values of the fermions are \[\delta_{\varepsilon,\bar\varepsilon} \langle \psi^Q \rangle \sim \varepsilon \langle F^Q \rangle ~, \qquad \delta_{\varepsilon,\bar\varepsilon} \langle \psi^{\tilde Q} \rangle \sim \varepsilon \langle F^{\tilde Q} \rangle ~, \qquad \delta_{\varepsilon,\bar\varepsilon} \langle \lambda \rangle \sim \varepsilon \langle D \rangle ~,\] hence we see explicitly that the vacuum is supersymmetric if and only if \(\langle F^Q \rangle = \langle F^{\tilde Q} \rangle = \langle D \rangle = 0\).

The stationary points of the potential solve \[\begin{split} \frac{\partial V}{\partial \phi^Q} & = \big(|m|^2 + g^2 (|\phi^Q|^2 - |\phi^{\tilde{Q}}|^2 - \xi)\big)\bar\phi_Q = 0 ~, \\ \frac{\partial V}{\partial \phi^{\tilde Q}} & = \big(|m|^2 - g^2 (|\phi^Q|^2 - |\phi^{\tilde{Q}}|^2 - \xi)\big)\bar\phi_{\tilde Q} = 0 ~. \end{split}\] Given that \(m\neq 0\), let us define \[U = \frac{g \phi^Q}{m} ~, \qquad \tilde U = \frac{g\phi^{\tilde Q}}{m} ~, \qquad X = \frac{g^2 \xi}{|m|^2} > 0 ~,\] so that these equations simplify to \[(1 + |U|^2 - |\tilde{U}|^2 - X)\bar U = 0 ~, \qquad (1 - |U|^2 + |\tilde{U}|^2 + X) \bar{\tilde{U}} = 0 ~.\] Depending on the value of \(X\), there are either one or two solutions to these equations. These are solution (i), \(U = \tilde U = 0\), which leads to the vacuum expectation values \[\begin{gathered} \langle \phi^Q \rangle = \langle \phi^{\tilde Q} \rangle = 0 ~, \qquad \langle F^Q \rangle = \langle F^{\tilde Q} \rangle = 0 ~, \qquad \langle Q \rangle = \langle \tilde{Q} \rangle = 0 ~, \\ \langle D \rangle = \frac{|m|^2}{g} X ~, \qquad \langle V\rangle = \frac{|m|^4}{2g^2} X^2 ~, \end{gathered}\] and, for \(X\geq 1\), solution (ii), \(|U| = \sqrt{X-1}\), \(\tilde U = 0\), which leads to the vacuum expectation values \[\begin{gathered} \langle \phi^Q \rangle = \frac{m e^{i\alpha(x)}}{g} \sqrt{X-1} ~, \qquad \langle \phi^{\tilde Q} \rangle = 0 ~, \qquad \langle F^Q \rangle = 0 ~, \qquad \langle F^{\tilde Q} \rangle = \frac{|m|^2 e^{-i\alpha(x)}}{g} \sqrt{X-1} ~, \\ \langle Q \rangle = \frac{m e^{i\alpha(x)}}{g} \sqrt{X-1} ~, \qquad \langle \tilde Q \rangle = -\theta\theta \frac{|m|^2 e^{-i\alpha(x)}}{g} \sqrt{X-1} ~, \\ \langle D \rangle = \frac{|m|^2}{g} ~, \qquad \langle V \rangle = \frac{|m|^4}{2g^2} (2X - 1) ~, \end{gathered}\] where the phase \(\alpha(x)\) parametrises the \(\mathrm{U}(1)\) gauge freedom. This is only a solution for \(X\geq 1\) since \(|U| \in \mathbb{R}\) and \(\sqrt{X-1}\) is imaginary for \(X<1\). For \(X=1\) solution (ii) coincides with solution (i). In principle there could be two other solutions to these equations \[\begin{gathered} U = 0 ~, \qquad |\tilde U| = \sqrt{-X-1} ~, \\ |U|^2 - |\tilde U|^2 = X-1 = X+1 ~. \end{gathered}\] The first of these is not a solution since \(|\tilde U| \in \mathbb{R}\) and \(X > 0\) implies that \(\sqrt{-X-1}\) is imaginary, while the second immediately leads to a contradiction. When computing the vacuum expectation values of \(Q\) and \(\tilde Q\) above, we have used that the vacuum expectation values of fermions vanish by the Poincaré invariance of the vacuum.

For \(X<1\) the only solution is solution (i), which is the global minimum. For \(X>1\) we have that \[(X-1)^2 > 0 \qquad \Rightarrow \qquad X^2 > 2X-1 ~,\] hence the global minimum is given by solution (ii), while solution (i) is a local maximum. For \(X=1\) the global minimum is given by both solutions, which coincide. It follows that for \(X\leq 1\) we have D-term spontaneous supersymmetry breaking since \(\langle F^Q \rangle = \langle F^{\tilde Q} \rangle = 0\) and \(\langle D \rangle \neq 0\), while for \(X > 1\) we have both F-term and D-term spontaneous supersymmetry breaking.

Using the results of exercise 6.2 to compute the component field expansion of the action, expanding around a general vacuum, and extracting those terms quadratic in the fields, we find \[\begin{split} \mathcal{S}_2 & = \int d^4x\,\big(-\frac{1}{4}F^{\mu\nu} F_{\mu\nu} + \frac{\theta g^2}{32\pi^2}\tilde F^{\mu\nu}F_{\mu\nu} \big) \\ & \quad + \int d^4x \, \big( (\partial_\mu \bar\phi_Q + i g \langle \bar\phi_Q\rangle A_\mu ) (\partial^\mu \phi^Q - i g \langle \phi^Q\rangle A^\mu) + (\partial_\mu \bar\phi_{\tilde Q} - i g A_\mu \langle \bar\phi_{\tilde Q}\rangle) (\partial^\mu \phi^{\tilde Q} + i g A^\mu \langle \phi^{\tilde Q}\rangle) \big) \\ & \quad - \int d^4 x \, \frac{g^2}{2} \begin{pmatrix} \phi^Q & \phi^{\tilde Q} & \bar\phi_Q & \bar\phi_{\tilde Q} \end{pmatrix} M_{\text{scalar}} \begin{pmatrix} \phi^Q & \phi^{\tilde Q} & \bar\phi_Q & \bar\phi_{\tilde Q} \end{pmatrix}^\mathrm{t} \\ & \quad + \int d^4 x\, \big( - i \bar\lambda \bar\sigma^\mu \partial_\mu\lambda -i \bar\psi_Q \bar\sigma^\mu \partial_\mu \psi^Q -i \bar\psi_{\tilde Q} \bar\sigma^\mu \partial_\mu \psi^{\tilde Q} \big) \\ & \quad + \int d^4 x\, \big( + i g \sqrt{2} \langle \bar\phi_Q\rangle \lambda \psi^Q - i g \sqrt{2} \langle \bar\phi_{\tilde Q}\rangle \lambda \psi^{\tilde Q} - i g \sqrt{2} \langle \phi^Q\rangle \bar \lambda \bar \psi_Q + i g \sqrt{2} \langle \phi^{\tilde Q}\rangle \bar \lambda \bar \psi_{\tilde Q} \\ & \hspace{300pt} - m \psi^Q \psi^{\tilde Q} - \bar m \bar\psi_Q \bar\psi_{\tilde Q} \big) ~, \end{split}\] where \[M_{\text{scalar}} = \small \begin{pmatrix} \langle \bar\phi_Q\rangle^2 & - \langle \bar\phi_Q\rangle\langle \bar\phi_{\tilde Q} \rangle & |\langle \phi^Q\rangle|^2 +\frac{|m|^2 - g\langle D\rangle}{g^2} & - \langle \bar\phi_Q\rangle\langle \phi^{\tilde Q} \rangle \\ - \langle \bar\phi_Q\rangle\langle \bar\phi_{\tilde Q} \rangle & \langle \bar\phi_{\tilde Q}\rangle^2 & - \langle \bar\phi_{\tilde Q}\rangle\langle \phi^{Q} \rangle & |\langle \phi^{\tilde Q}\rangle|^2 +\frac{|m|^2 + g\langle D\rangle}{g^2} \\ |\langle \phi^Q\rangle|^2 +\frac{|m|^2 - g\langle D\rangle}{g^2} & - \langle \bar\phi_{\tilde Q}\rangle\langle \phi^{Q} \rangle & \langle \phi^Q\rangle^2 & - \langle \phi^Q\rangle \langle \phi^{\tilde Q}\rangle \\ - \langle \bar\phi_Q\rangle\langle \phi^{\tilde Q} \rangle & |\langle \phi^{\tilde Q}\rangle|^2 +\frac{|m|^2 + g\langle D\rangle}{g^2} & - \langle \phi^Q\rangle \langle \phi^{\tilde Q}\rangle & \langle \phi^{\tilde Q}\rangle^2 \end{pmatrix} \normalsize ~.\]

For solution (i), i.e. \(X\leq 1\), we have \(\langle \phi^Q \rangle = \langle \phi^{\tilde{Q}} \rangle = 0\) and the quadratic action simplifies to \[\begin{split} \mathcal{S}_2 & = \int d^4x\,\big(-\frac{1}{4}F^{\mu\nu} F_{\mu\nu} + \frac{\theta g^2}{32\pi^2}\tilde F^{\mu\nu}F_{\mu\nu} + \partial_\mu \bar\phi_Q \partial^\mu \phi^Q + \partial_\mu \bar\phi_{\tilde Q}\partial^\mu \phi^{\tilde Q} \big) \\ & \quad + \int d^4x \, \big( - i \bar\lambda \bar\sigma^\mu \partial_\mu\lambda -i \bar\psi_Q \bar\sigma^\mu \partial_\mu \psi^Q -i \bar\psi_{\tilde Q} \bar\sigma^\mu \partial_\mu \psi^{\tilde Q} - m \psi^Q \psi^{\tilde Q} - \bar m \bar\psi_Q \bar\psi_{\tilde Q} \big) ~. \end{split}\] Therefore, all the fields are massless apart from \(\psi^Q\) and \(\psi^{\tilde Q}\), which together form a Dirac fermion of mass \(|m|\). The gaugino \(\lambda\) is massless, hence it is the goldstino.

For solution (ii), i.e. \(X>1\), we have \(\langle \phi^{\tilde{Q}} \rangle = 0\), \(\langle D \rangle = \frac{|m|^2}{g}\) and we use the \(\mathrm{U}(1)\) gauge symmetry to fix \(\phi^Q = \bar\phi_Q = \frac{1}{\sqrt{2}}\varphi \in \mathbb{R}\), hence \(\langle \phi^Q \rangle = \langle \bar\phi_Q\rangle = \frac{|m|}{g} \sqrt{X-1} \in \mathbb{R}\). The quadratic action then becomes \[\begin{split} \mathcal{S}_2 & = \int d^4x\,\big(-\frac{1}{4}F^{\mu\nu} F_{\mu\nu} + \frac{\theta g^2}{32\pi^2}\tilde F^{\mu\nu}F_{\mu\nu} +\frac{1}{2}\partial_\mu \varphi \partial^\mu \varphi + \partial_\mu \bar\phi_{\tilde Q}\partial^\mu \phi^{\tilde Q} \big) \\ & \quad + \int d^4 x\, \big( |m|^2(X-1) A_\mu A^\mu - |m|^2 (X-1) \varphi^2 - 2|m|^2\bar\phi_{\tilde Q}\phi^{\tilde Q}\big) \\ & \quad + \int d^4x \, \big( - i \bar\lambda \bar\sigma^\mu \partial_\mu\lambda -i \bar\psi_Q \bar\sigma^\mu \partial_\mu \psi^Q -i \bar\psi_{\tilde Q} \bar\sigma^\mu \partial_\mu \psi^{\tilde Q} \big) \\ & \quad + \int d^4 x\, \big( - \psi^Q (m \psi^{\tilde{Q}} - i |m| \sqrt{2} \sqrt{X-1} \lambda ) - \bar\psi_Q (\bar m \bar\psi_{\tilde Q} + i |m| \sqrt{2} \sqrt{X-1} \bar \lambda) \big) ~. \end{split}\] To diagonalise the fermionic mass terms in this quadratic action we define \[\psi^{\tilde Q} = \frac{\sqrt{\frac{\bar m}{m}} \psi^{\hat Q} +\sqrt{2}\sqrt{X-1} \psi_{\text{G}}}{\sqrt{2X-1}} ~, \qquad \lambda = \frac{i(\sqrt{2}\sqrt{X-1}\psi^{\hat Q} - \sqrt{\frac{m}{\bar m}} \psi_{\text{G}})}{\sqrt{2X-1}}~,\] so that \[\begin{split} \mathcal{S}_2 & = \int d^4x\,\big(-\frac{1}{4}F^{\mu\nu} F_{\mu\nu} + \frac{\theta g^2}{32\pi^2}\tilde F^{\mu\nu}F_{\mu\nu} +\frac{1}{2}\partial_\mu \varphi \partial^\mu \varphi + \partial_\mu \bar\phi_{\tilde Q}\partial^\mu \phi^{\tilde Q} \big) \\ & \quad + \int d^4 x\, \big( |m|^2(X-1) A_\mu A^\mu - |m|^2 (X-1) \varphi^2 - 2|m|^2\bar\phi_{\tilde Q}\phi^{\tilde Q}\big) \\ & \quad + \int d^4x \, \big( - i \bar\psi_{\text{G}} \bar\sigma^\mu \partial_\mu\psi_{\text{G}} -i \bar\psi_Q \bar\sigma^\mu \partial_\mu \psi^Q -i \bar\psi_{\hat Q} \bar\sigma^\mu \partial_\mu \psi^{\hat Q} \big) \\ & \quad + \int d^4 x\, \big( - |m|\sqrt{2X-1} \psi^Q \psi^{\hat Q} - |m| \sqrt{2X-1} \bar\psi_Q \bar \psi_{\hat Q} \big) ~. \end{split}\] This action describes a massive vector \(A_\mu\) of mass \(\sqrt{2}|m|\sqrt{X-1}\), a massive real scalar of mass \(\sqrt{2}|m|\sqrt{X-1}\) and a massive complex scalar of mass \(\sqrt{2}|m|\). Here, the vector \(A_\mu\) has acquired its mass via the Higgs mechanism. Together \(\psi^Q\) and \(\psi^{\hat Q}\) form a Dirac fermion of mass \(|m|\sqrt{2X-1}\), while \(\psi_{\text{G}}\) is the massless goldstino.

For tree-level spontaneous supersymmetry breaking in a theory with canonical kinetic terms, the mass spectrum of bosons and fermions satisfies the sum rule \[\mathop{\mathrm{str}}\big(M^2\big) = \mathop{\mathrm{tr}}\big((-1)^F M \big) = \sum_s (-1)^{2s} (2s+1) m_s^2 = -2 \sum_I \sum_A g_A \langle D_A \rangle \mathop{\mathrm{Tr}}(T^A_{{\scriptscriptstyle{R}}_I}) ~,\] where \(M^2\) is the mass-squared matrix, \(s\) is the spin and \(m_s^2\) are the masses of the spin \(s\) particles. The factor of \(2s+1\) accounts for the degeneracy of massive particles of spin \(s\).

This is derived as follows. We start by defining \[\begin{aligned} \bar F_{IJ} & = \frac{\partial \bar F_J}{\partial\Phi^I} = \frac{\partial\bar F_I }{\partial\Phi^J} = \partial_I \partial_J W ~, \qquad & F^{IJ} & = \frac{\partial F^J }{\partial\bar \Phi_I} = \frac{\partial F^I }{\partial\bar\Phi_J}= \bar\partial^I \bar\partial^J \bar W ~, \\ \bar D_{AI} & = \frac{\partial D_A}{\partial\Phi^I }= - g_A \bar\phi_I T^A_{{\scriptscriptstyle{R}}_I} ~, \qquad & D^I_A & = \frac{\partial D_A }{\partial\bar\Phi_I} = - g_A T^A_{{\scriptscriptstyle{R}}_I} \phi^I ~. \end{aligned}\] Mass terms for the vectors come from the terms \(\sum_I \bar D_{{\scriptscriptstyle{R}}\mu} \bar\phi_I D_{\scriptscriptstyle{R}}^\mu \phi^I\) in the action. In particular, we find the mass terms \[\sum_I \sum_{A,B} g_A g_B \langle \bar\phi_I \rangle T^A_{{\scriptscriptstyle{R}}_I} T^B_{{\scriptscriptstyle{R}}_I} \langle \phi^I \rangle A_{\mu A} A^{\mu}_B ~.\] Therefore, the mass-squared matrix for the vectors is \[(M_1^2)_{AB} = 2 \sum_I \langle \bar D_{AI} \rangle \langle D^I_B \rangle ~.\] The mass terms for the fermions are \[-\frac{1}{2} \sum_{I,J}\sum_{A,B}\begin{pmatrix} \psi^I & \lambda_B \end{pmatrix} M_{\frac{1}{2}} \begin{pmatrix} \psi^J \\ \lambda_A \end{pmatrix} ~, \qquad M_{\frac{1}{2}} = \begin{pmatrix} \langle \bar F_{IJ} \rangle & \langle \sqrt{2}i\bar D_{BI} \rangle \\ \sqrt{2}i\langle \bar D_{AJ} \rangle & 0 \end{pmatrix} ~,\] where \(M_{\frac{1}{2}}\) is the complex mass matrix. Therefore, the mass-squared matrix for the fermions is \[M_{\frac{1}{2}} M_{\frac{1}{2}}^\dagger = \begin{pmatrix} \sum_K \langle \bar F_{IK} \rangle \langle F^{JK} \rangle + 2 \sum_C \langle \bar D_{CI} \rangle \langle D_{C}^J \rangle & -\sqrt{2}i \sum_K \langle \bar F_{IK}\rangle \langle D_B^K \rangle \\ \sqrt{2}i \sum_K \langle \bar D_{AK}\rangle \langle F^{JK} \rangle & 0 \end{pmatrix} ~.\] The complex mass-squared matrix for the scalars is \[M_0^2 = \begin{pmatrix} \langle \partial_I \bar\partial^K V \rangle & \langle \partial_I \partial_L V \rangle \\ \langle \bar\partial^J \bar\partial^L V \rangle & \langle \bar\partial^J \partial_L V \rangle \end{pmatrix} ~.\] Using that \(V = \sum_I \bar F_I F^I + \frac{1}{2} \sum_A D_A D_A\), we find \[\begin{split} M_0^2 & = \left( \begin{array}{c} \sum_M \langle \bar F_{IM}\rangle \langle F^{KM} \rangle + \sum_A \langle D_A^K \rangle \langle \bar D_{AI}\rangle + \sum_A \langle D_A\rangle D_A{}_I^K \\ \sum_M \langle \bar F_M \rangle \langle F^{JKM} \rangle + \sum_A \langle D_A^J \rangle \langle D_A^K \rangle \end{array} \right. \\ & \hspace{125pt} \left. \begin{array}{c} \sum_M \langle F^M \rangle \langle \bar F_{ILM} \rangle + \sum_A \langle \bar D_{AI} \rangle \langle \bar D_{AL} \rangle \\ \sum_M \langle \bar F_{LM} \rangle \langle F^{JM} \rangle + \sum_A \langle D^J_A \rangle \langle \bar D_{AL} \rangle + \sum_A \langle D_A \rangle D_A{}_L^J \end{array}\right) ~, \end{split}\] where \(D_A{}_J^I = \bar\partial^I \partial_J D_A = -g_A T^A_{{\scriptscriptstyle{R}}_I} \delta_J^I\), \(\bar F_{IJK} = \partial_I \partial_J \partial_K W\) and \(F^{IJK} = \bar\partial^I \bar\partial^J \bar\partial^K \bar W\).

Therefore, \[\begin{split} \mathop{\mathrm{tr}}(M_1^2) & = 2 \sum_I \sum_A \langle \bar D_{AI} \rangle \langle D_A^I \rangle ~, \\ \mathop{\mathrm{tr}}(M_{\frac{1}{2}} M_{\frac{1}{2}}^\dagger) & = \sum_{I,J} \langle \bar F_{IJ} \rangle \langle F^{IJ} \rangle + 2 \sum_I \sum_A \langle \bar D_{AI} \rangle \langle D_{A}^I \rangle ~, \\ \mathop{\mathrm{tr}}(M_0^2) & = 2\sum_{I,J} \langle \bar F_{IJ}\rangle \langle F^{IJ} \rangle + 2 \sum_I \sum_A \langle D_A^I \rangle \langle \bar D_{AI}\rangle -2 \sum_I \sum_A g_A \langle D_A \rangle \mathop{\mathrm{Tr}}(T^A_{{\scriptscriptstyle{R}}_I}) ~, \end{split}\] and \[\mathop{\mathrm{str}}(M^2) = 3 \mathop{\mathrm{tr}}(M_1^2) - 2\mathop{\mathrm{tr}}(M_{\frac{1}{2}} M_{\frac{1}{2}}^\dagger) + \mathop{\mathrm{tr}}(M_0^2) = -2 \sum_I \sum_A g_A \langle D_A \rangle \mathop{\mathrm{Tr}}(T^A_{{\scriptscriptstyle{R}}_I}) ~.\] If there are no \(\mathrm{U}(1)\) factors of \(\mathrm{G}\) then we have that \(\mathop{\mathrm{str}}(M^2) = 0\) since for compact simple Lie groups \(\mathop{\mathrm{Tr}}(T^a_{{\scriptscriptstyle{R}}_I}) = 0\). We also find that \(\mathop{\mathrm{str}}(M^2) = 0\) for F-term spontaneous supersymmetry breaking. Another consequence of this tree-level analysis is that if the gauge symmetry is not spontaneously broken then \(M_1^2 = 0\), which implies that \(\langle \bar D_{AI} \rangle = \langle D_{AI} \rangle = 0\), hence the gauginos are massless. This is ruled out by experiment. Similarly, for the MSSM one finds that \(\mathop{\mathrm{str}}(M^2) = 0\), which is also not compatible with observations. Therefore, since the supertrace mass formula holds at tree level, it follows that spontaneous supersymmetry breaking cannot occur at tree level in the MSSM. Eventually, one is led to consider the spontaneous breaking of supersymmetry by non-perturbative effects.

The factor of \(i\) is a matter of convention. We are following the conventions more often used in physics, as opposed to those used in mathematics.↩︎

Here we have used that \(N_\alpha{}^\gamma N_\beta{}^\delta \epsilon_{\gamma\delta}\) is a \(2 \times 2\) antisymmetric matrix and thus is proportional to \(\epsilon_{\alpha\beta}\). The proportionality constant can be fixed by contracting both sides with \(\epsilon^{\alpha\beta}\).↩︎

This is a consistent set of conventions since \(\psi_\alpha = \epsilon_{\alpha\beta}\psi^\beta = \epsilon_{\alpha\beta}\epsilon^{\beta\gamma} \psi_\gamma = \psi_\alpha\), etc.↩︎

For superconformal theories the R-symmetry is unavoidably part of the superalgebra since it appears in the anticommutation relations of the Poincaré supercharges with the superconformal charges.↩︎

This is referred to as the Haag-Łopuszański-Sohnius theorem.↩︎

This subalgebra is the supersymmetric analogue of the algebra of the little group introduced in subsection 2.4.↩︎

In section 4 we will see an off-shell generalisation using superfields.↩︎

In the massless case, those supermultiplets relevant for SQFT have \(|\lambda|\leq1\) for all states, while those relevant for supergravity have \(|\lambda|\leq2\).↩︎

The gravitino is usually only considered when the theory also contains a graviton.↩︎

In the massive case, those supermultiplets relevant for SQFT have \(|s_z|\leq1\) for all states.↩︎

For even \(\mathcal{N}\) we have creation operators \((a_\alpha^r)^\dagger\) and \((b_\alpha^r)^\dagger\) giving \(\frac{\mathcal{N}}{2} \times 2 \times 2 = 2\mathcal{N}\) in total. For odd \(\mathcal{N}\) we have an additional pair associated with the last supercharge \(Q_\alpha^{\mathcal{N}}\).↩︎

The BPS bound is named for Bogomol’nyi, Prasad and Sommerfeld, who found a related bound for solitons in non-supersymmetric theories. That this bound follows from the supersymmetry algebra was discovered by Olive and Witten.↩︎

In terms of \(\mathcal{N}=1\) supermultiplets, the \(\mathcal{N}=1\) massless vector multiplet combines with the \(\mathcal{N}=1\) massless chiral multiplet (in the adjoint representation of the gauge group) that is part of the same \(\mathcal{N}=2\) massless vector multiplet.↩︎

In terms of \(\mathcal{N}=2\) supermultiplets, the \(\mathcal{N}=2\) massless vector multiplet combines with the \(\mathcal{N}=2\) massless hypermultiplet (in the adjoint representation of the gauge group) that is part of the same \(\mathcal{N}=4\) massless vector multiplet.↩︎

Note that when both approaches are available they are equivalent. As an analogy, consider \(\mathrm{U}(1)\) Yang-Mills theory. We can either write the action in terms of the components \(A_0\) and \(A_i\), or in terms of \(A_\mu\). Writing in terms of components highlights the distinction between time and space, while writing in terms of \(A_\mu\) makes Lorentz invariance manifest.↩︎

When we consider differential operators in commutators or anticommutators such as \([P_\mu,\phi(x)] = [-i \partial_\mu,\phi(x)]\), the differential operator acts on everything to the right. This means that \([-i \partial_\mu,\phi(x)] F(x) = -i \partial_\mu\phi(x) F(x)\) for all \(F(x)\).↩︎

This is analogous to constraining a Dirac spinor, a reducible representation of the Lorentz algebra, to be either a Majorana or Weyl spinor, which are irreducible representations of the Lorentz algebra.↩︎

A non-vanishing contribution could only come from \(\bar\theta^{\dot\alpha}\bar\theta^{\dot\beta}\bar\theta^{\dot\gamma}\), which is itself zero since \(\bar\theta^{\dot\alpha}\) is anticommuting and the index \(\dot\alpha=1,2\).↩︎